Practice Final, Fall 2024

The integral is $\int_{-1}^{1} (5x^3 - 2x + \arctan x) \, dx$.
The interval of integration is $[-1, 1]$. This is a symmetric interval because it is of the form $[-a, a]$ with $a=1$.
The integrand is $f(x) = 5x^3 - 2x + \arctan x$.


When integrating over a symmetric interval $[-a, a]$, it's useful to check if the integrand $f(x)$ is an odd function or an even function.
- Recall: A function $f$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain.
- Recall: A function $f$ is even if $f(-x) = f(x)$ for all $x$ in its domain.

Let's check the symmetry of $f(x) = 5x^3 - 2x + \arctan x$ by replacing $x$ with $-x$:
$$f(-x)=5(-x{)}^3-2(-x)+\arctan (-x)$$
f(-x) = 5(-x)^3 - 2(-x) + \arctan(-x)

We know that:
- $(-x)^3 = -x^3$
- $\arctan(-x) = -\arctan(x)$ (Arctangent is an odd function)
Substitute these back:
$$\begin{align*}$$
f(-x) &= 5(-x^3) - 2(-x) + (-\arctan x) \\
&= -5x^3 + 2x - \arctan x
\end{align*}
Now, compare this to $-f(x)$:
$$\begin{align*}$$
-f(x) &= -(5x^3 - 2x + \arctan x) \\
&= -5x^3 + 2x - \arctan x
\end{align*}
Since $f(-x) = -f(x)$, the integrand $f(x) = 5x^3 - 2x + \arctan x$ is an odd function.
Alternatively, you could note that $5x^3$ is odd, $-2x$ is odd, and $\arctan x$ is odd. The sum of odd functions is always an odd function.


There is a property for definite integrals over symmetric intervals:
If $f(x)$ is an odd function and continuous on $[-a, a]$, then $\int_{-a}^{a} f(x) \, dx = 0$.
Geometrically, this is because the signed area from $-a$ to 0 cancels out the signed area from 0 to $a$.

Since our integrand $f(x) = 5x^3 - 2x + \arctan x$ is an odd function and the interval is $[-1, 1]$, we can directly apply this property.


$$\int_{-1}^{1} (5x^3 - 2x + \arctan x) \, dx = 0$$
The value of the definite integral is 0. This corresponds to option (A).

$$\boxed{{\displaystyle 0}}$$
\boxed{ 0 }

The definition of the definite integral as a limit of Riemann sums is:
$${\int }_a^{b}f(x)\hspace{0.17em}dx=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^{n}f(x_i^{\ast })\mathrm{\Delta }x$$
\int_{a}^{b} f(x) \, dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*) \Delta x

Where:
- $[a, b]$ is the interval of integration.
- $n$ is the number of subintervals.
- $\Delta x = \frac{b-a}{n}$ is the width of each subinterval.
- $x_i^*$ is the sample point in the $i$-th subinterval.


The given limit expression is:
$$L=\underset{n\to \mathrm{\infty }}{\lim }\sum _{i=1}^n\frac{1}{1+(i\mathrm{/}n{)}^2}\cdot \frac{1}{n}$$
L = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{1+(i/n)^2} \cdot \frac{1}{n}

We need to match this structure $\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$ by identifying each component. The summation is already written using sigma notation.


The term multiplying the summation terms inside the limit is $\frac{1}{n}$. This strongly suggests that $\Delta x$ corresponds to this term:
$$\Delta x = \frac{1}{n}$$
Since $\Delta x = \frac{b-a}{n}$, equating these implies the length of the integration interval is $b-a = 1$.
A standard and simple choice for an interval of length 1 is $[a, b] = [0, 1]$.


Assuming the interval is $[0, 1]$ and $\Delta x = 1/n$, let's consider the most common choice for sample points: the right endpoints.
The formula for right endpoints is $x_i = a + i \Delta x$. With $a=0$ and $\Delta x = 1/n$:
$$x_i = 0 + i\left(\frac{1}{n}\right) = \frac{i}{n}$$
Now we look at the term inside the summation in the given limit: $\frac{1}{1+(i/n)^2}$.
If we substitute our right endpoint $x_i = i/n$ into this term, we get:
$$\frac{1}{1+(x_i)^2}$$
This perfectly matches the form $f(x_i^*)$ in the Riemann sum definition if we let $x_i^* = x_i$ (right endpoints) and define the function $f(x)$ as:
$$f(x) = \frac{1}{1+x^2}$$
The structure $\sum f(x_i) \Delta x$ aligns perfectly with the given sum $\sum_{i=1}^{n} \frac{1}{1+(i/n)^2} \cdot \frac{1}{n}$ if we choose $a=0, b=1, \Delta x = 1/n, x_i = i/n$, and $f(x) = 1/(1+x^2)$.


Using the identified components:
- Interval: $[a, b] = [0, 1]$
- Function: $f(x) = \frac{1}{1+x^2}$
The limit of the Riemann sum corresponds to the definite integral:
$${\int }_a^{b}f(x)\hspace{0.17em}dx={\int }_0^1\frac{1}{1+x^2}\hspace{0.17em}dx$$
\int_{a}^{b} f(x) \, dx = \int_{0}^{1} \frac{1}{1+x^2} \, dx



The given limit corresponds to the definite integral $\int_{0}^{1} \frac{1}{1+x^2} \, dx$.

The series is $S = \sum_{n=1}^{\infty} \frac{5^n + 2^n}{6^n}$.
Let $a_n$ be the general term. We can split the fraction:
$$\begin{align*} a_n &= \frac{5^n + 2^n}{6^n} \\ &= \frac{5^n}{6^n} + \frac{2^n}{6^n} \\ &= \left(\frac{5}{6}\right)^n + \left(\frac{2}{6}\right)^n \\ &= \left(\frac{5}{6}\right)^n + \left(\frac{1}{3}\right)^n \end{align*}$$


Since the series $\sum a_n$ is a sum of two terms, we can investigate the convergence of the series corresponding to each term. If both converge, the sum of the original series is the sum of the individual series.
Property of Convergent Series: If $\sum u_n$ and $\sum v_n$ both converge, then $\sum (u_n + v_n) = \sum u_n + \sum v_n$.
Let's split the series:
$$\begin{align*} S &= \sum_{n=1}^{\infty} \left[ \left(\frac{5}{6}\right)^n + \left(\frac{1}{3}\right)^n \right] \\ &= \sum_{n=1}^{\infty} \left(\frac{5}{6}\right)^n + \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n \end{align*}$$


The first series is $S_1 = \sum_{n=1}^{\infty} \left(\frac{5}{6}\right)^n$.
This is a geometric series.
- The common ratio is $r_1 = 5/6$.
- The first term (when $n=1$) is $a_1 = (5/6)^1 = 5/6$.
Since the absolute value of the common ratio $|r_1| = 5/6 < 1$, the series converges.
The sum of a convergent geometric series starting from $n=1$ is given by $\frac{\text{first term}}{1 - r}$.
$$\begin{align*} S_1 &= \frac{a_1}{1 - r_1} \\ &= \frac{5/6}{1 - 5/6} \\ &= \frac{5/6}{1/6} \\ &= 5 \end{align*}$$


The second series is $S_2 = \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n$.
This is also a geometric series.
- The common ratio is $r_2 = 1/3$.
- The first term (when $n=1$) is $a_2 = (1/3)^1 = 1/3$.
Since the absolute value of the common ratio $|r_2| = 1/3 < 1$, the series converges.
The sum is:
$$\begin{align*} S_2 &= \frac{a_2}{1 - r_2} \\ &= \frac{1/3}{1 - 1/3} \\ &= \frac{1/3}{2/3} \\ &= \frac{1}{2} \end{align*}$$
Both series converge, so we can find the sum of the original series by adding the sums $S_1$ and $S_2$.


The sum of the original series is $S = S_1 + S_2$.
$$\begin{align*}$$
S &= 5 + \frac{1}{2} \\
&= \frac{10}{2} + \frac{1}{2} \\
&= \frac{11}{2}
\end{align*}


The sum of the infinite series is $\frac{11}{2}$.
Comparing this to the options:
(A) 6
(B) 11/2
(C) 7/2
(D) The series is divergent

The correct option is (B).
$$\boxed{{\displaystyle (B)\hspace{0.17em}\frac{11}{2}}}$$
\boxed{ (B) \, \frac{11}{2} }

The given series is $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n 3^n}$. This is a power series centered at $a=2$.
To find the radius of convergence, the Ratio Test is generally the most effective method.


Let $a_n = \frac{(x-2)^n}{n 3^n}$. The Ratio Test examines the limit:
$$L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$$
First, write out $a_{n+1}$:
$$a_{n+1} = \frac{(x-2)^{n+1}}{(n+1) 3^{n+1}}$$
Now form the ratio $\left| \frac{a_{n+1}}{a_n} \right|$ and simplify:
$$\begin{align*} \left| \frac{a_{n+1}}{a_n} \right| &= \left| \frac{(x-2)^{n+1}}{(n+1) 3^{n+1}} \div \frac{(x-2)^n}{n 3^n} \right| \\ &= \left| \frac{(x-2)^{n+1}}{(n+1) 3^{n+1}} \cdot \frac{n 3^n}{(x-2)^n} \right| \\ &= \left| \frac{(x-2)^{n+1}}{(x-2)^n} \cdot \frac{3^n}{3^{n+1}} \cdot \frac{n}{n+1} \right| \\ &= \left| (x-2) \cdot \frac{1}{3} \cdot \frac{n}{n+1} \right| \\ &= \frac{|x-2|}{3} \cdot \frac{n}{n+1} \end{align*}$$
(Since $n \ge 1$, $\frac{n}{n+1}$ is positive).


Now find the limit of the ratio as $n \to \infty$:
$$\begin{align*} L &= \lim_{n\to\infty} \left( \frac{|x-2|}{3} \cdot \frac{n}{n+1} \right) \end{align*}$$
Since $\frac{|x-2|}{3}$ does not depend on $n$, we can pull it out of the limit:
$$\begin{align*} L &= \frac{|x-2|}{3} \lim_{n\to\infty} \left( \frac{n}{n+1} \right) \end{align*}$$
Evaluate the remaining limit:
$$\begin{align*} \lim_{n\to\infty} \left( \frac{n}{n+1} \right) &= \lim_{n\to\infty} \left( \frac{n/n}{(n+1)/n} \right) \\ &= \lim_{n\to\infty} \left( \frac{1}{1 + 1/n} \right) \\ &= \frac{1}{1 + 0} = 1 \end{align*}$$
Substitute this back into the expression for $L$:
$$L = \frac{|x-2|}{3} \cdot 1 = \frac{|x-2|}{3}$$


The Ratio Test states that the series converges absolutely if $L < 1$.
$$\frac{|x-2|}{3} < 1$$
Solve this inequality for $|x-2|$:
$$|x-2| < 3$$
This inequality defines the interval of convergence, centered at $a=2$. The form is $|x-a| < R$, where $R$ is the radius of convergence.


Comparing $|x-2| < 3$ to the standard form $|x-a| < R$, we see that the radius of convergence is $R = 3$.


The radius of convergence for the series $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n 3^n}$ is 3.

$$\boxed{{\displaystyle 3}}$$
\boxed{ 3 }

Integral: $I_A = \int_{1}^{\infty} \frac{\ln x}{x} dx$
Type: This is an improper integral of Type 1 due to the infinite upper limit.
Method: We can try to evaluate it using a limit, possibly with u-substitution.
Let $u = \ln x$. Then $du = \frac{1}{x} dx$.
Change limits:
- When $x=1$, $u = \ln 1 = 0$.
- As $x \to \infty$, $u = \ln x \to \infty$.
Substitute:
$$\begin{align*} I_A &= \int_{0}^{\infty} u \, du \\ &= \lim_{b\to\infty} \int_{0}^{b} u \, du \\ &= \lim_{b\to\infty} \left[ \frac{u^2}{2} \right]_{0}^{b} \\ &= \lim_{b\to\infty} \left( \frac{b^2}{2} - \frac{0^2}{2} \right) \\ &= \lim_{b\to\infty} \frac{b^2}{2} \\ &= \infty \end{align*}$$
Conclusion: The integral $\int_{1}^{\infty} \frac{\ln x}{x} dx$ diverges.


Integral: $I_B = \int_{0}^{1} \frac{1}{\sqrt[3]{x}} dx$
Type: This is an improper integral of Type 2 because the integrand $\frac{1}{x^{1/3}}$ has an infinite discontinuity (vertical asymptote) at the lower limit $x=0$.
Method: This fits the form of a p-integral at a finite endpoint: $\int_{0}^{a} \frac{1}{x^p} dx$.
Rewrite the integrand: $\frac{1}{\sqrt[3]{x}} = \frac{1}{x^{1/3}}$.
Here, $p = 1/3$.
Recall the convergence rule for Type 2 p-integrals: $\int_{0}^{a} \frac{1}{x^p} dx$ converges if $p < 1$ and diverges if $p \ge 1$.
Since $p = 1/3 < 1$, the integral converges.
Conclusion: The integral $\int_{0}^{1} \frac{1}{\sqrt[3]{x}} dx$ converges.


Integral: $I_C = \int_{1}^{\infty} \frac{x}{x^2+1} dx$
Type: This is an improper integral of Type 1 due to the infinite upper limit.
Method 1: Evaluation using u-substitution.
Let $u = x^2+1$. Then $du = 2x \, dx$, so $x \, dx = \frac{1}{2} du$.
Change limits:
- When $x=1$, $u = 1^2+1 = 2$.
- As $x \to \infty$, $u = x^2+1 \to \infty$.
Substitute:
$$\begin{align*} I_C &= \int_{2}^{\infty} \frac{1}{u} \left( \frac{1}{2} du \right) \\ &= \frac{1}{2} \int_{2}^{\infty} \frac{1}{u} \, du \\ &= \frac{1}{2} \lim_{b\to\infty} \int_{2}^{b} \frac{1}{u} \, du \\ &= \frac{1}{2} \lim_{b\to\infty} [\ln|u|]_{2}^{b} \\ &= \frac{1}{2} \lim_{b\to\infty} (\ln b - \ln 2) \\ &= \frac{1}{2} (\infty - \ln 2) \\ &= \infty \end{align*}$$
Conclusion (Method 1): The integral $\int_{1}^{\infty} \frac{x}{x^2+1} dx$ diverges.
Method 2: Limit Comparison Test (LCT).
Compare $a_n = \frac{x}{x^2+1}$ with $b_n = \frac{x}{x^2} = \frac{1}{x}$.
$$\begin{align*} \lim_{x\to\infty} \frac{a_n}{b_n} &= \lim_{x\to\infty} \frac{x/(x^2+1)}{1/x} \\ &= \lim_{x\to\infty} \frac{x^2}{x^2+1} \\ &= \lim_{x\to\infty} \frac{1}{1+1/x^2} \\ &= \frac{1}{1+0} \\ &= 1 \end{align*}$$
Since the limit is finite and positive (1), and $\int_{1}^{\infty} \frac{1}{x} dx$ diverges (p-integral with $p=1$), $I_C$ also diverges by LCT.


Integral: $I_D = \int_{0}^{\pi/2} \tan x \, dx$
Type: This is an improper integral of Type 2 because $\tan x = \frac{\sin x}{\cos x}$ has an infinite discontinuity (vertical asymptote) at the upper limit $x=\pi/2$ (where $\cos x = 0$).
Method: Evaluate using the limit definition.
The antiderivative of $\tan x$ is $-\ln|\cos x|$.
$$\begin{align*} I_D &= \lim_{b\to (\pi/2)^{-}} \int_{0}^{b} \tan x \, dx \\ &= \lim_{b\to (\pi/2)^{-}} [-\ln|\cos x|]_{0}^{b} \\ &= \lim_{b\to (\pi/2)^{-}} (-\ln|\cos b|) - (-\ln|\cos 0|) \\ &= \lim_{b\to (\pi/2)^{-}} (-\ln|\cos b|) + \ln|1| \\ &= \lim_{b\to (\pi/2)^{-}} (-\ln|\cos b|) \end{align*}$$
As $b \to (\pi/2)^{-}$, $\cos b \to 0^{+}$. Therefore, $|\cos b| \to 0^{+}$.
As the argument of $\ln$ approaches $0^{+}$, $\ln(|\cos b|) \to -\infty$.
So, the limit is $-(-\infty) = +\infty$.
Conclusion: The integral $\int_{0}^{\pi/2} \tan x \, dx$ diverges.


Comparing the results for the four options:
(A) Diverges
(B) Converges
(C) Diverges
(D) Diverges

The only convergent integral is option (B).
$$\boxed{{\displaystyle (B)\hspace{0.17em}{\int }_0^1\frac{1}{\sqrt[3]{x}}dx}}$$
\boxed{ (B) \, \int_{0}^{1} \frac{1}{\sqrt[3]{x}} dx }

We need to find $f'(x) = \frac{d}{dx} \left( \int_{h(x)}^{g(x)} F(t) \, dt \right)$.
The Fundamental Theorem of Calculus, Part 1 (FTC Part 1), extended for variable limits of integration, states:

$$\frac{d}{dx}({\int }_{h(x)}^{g(x)}F(t)\hspace{0.17em}dt)=F(g(x))\cdot g^{\mathrm{\prime }}(x)-F(h(x))\cdot h^{\mathrm{\prime }}(x)$$

\frac{d}{dx} \left( \int_{h(x)}^{g(x)} F(t) \, dt \right) = F(g(x)) \cdot g'(x) - F(h(x)) \cdot h'(x)




This formula combines FTC Part 1 ($\frac{d}{dx} \int_a^x F(t) dt = F(x)$) with the Chain Rule. You evaluate the integrand at the upper limit and multiply by the upper limit's derivative, then subtract the integrand evaluated at the lower limit multiplied by the lower limit's derivative.


In our problem $f(x) = \int_{x^2}^{x^3} \cos(t^2) \, dt$:
- The integrand is $F(t) = \cos(t^2)$.
- The upper limit of integration is $g(x) = x^3$.
- The lower limit of integration is $h(x) = x^2$.


We need the derivatives of the upper and lower limits with respect to $x$:
$$g'(x) = \frac{d}{dx}(x^3) = 3x^2$$
$$h'(x) = \frac{d}{dx}(x^2) = 2x$$


We need to evaluate the integrand $F(t) = \cos(t^2)$ at the upper limit $g(x) = x^3$ and the lower limit $h(x) = x^2$:
$$F(g(x)) = F(x^3) = \cos((x^3)^2) = \cos(x^6)$$
$$F(h(x)) = F(x^2) = \cos((x^2)^2) = \cos(x^4)$$


Now substitute all the pieces into the formula: $f'(x) = F(g(x)) g'(x) - F(h(x)) h'(x)$
$$\begin{align*} f'(x) &= (\cos(x^6)) \cdot (3x^2) - (\cos(x^4)) \cdot (2x) \\ &= 3x^2 \cos(x^6) - 2x \cos(x^4) \end{align*}$$


The derivative is $3x^2 \cos(x^6) - 2x \cos(x^4)$.
Comparing this to the options:
(A) $\cos(x^6) - \cos(x^4)$
(B) $3x^2 \cos(x^6) - 2x \cos(x^4)$
(C) $\sin(x^6)(3x^2) - \sin(x^4)(2x)$
(D) $3x^2 \sin(x^6) - 2x \sin(x^4)$

The correct option is (B).
$$\boxed{{\displaystyle (B)\hspace{0.17em}3x^2\cos (x^6)-2x\cos (x^4)}}$$
\boxed{ (B) \, 3x^2 \cos(x^6) - 2x \cos(x^4) }

The average value of a continuous function $f(x)$ over a closed interval $[a, b]$ is given by the formula:
$$f_{avg}=\frac{1}{b-a}{\int }_a^{b}f(x)\hspace{0.17em}dx$$
f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx



- The function is $f(x) = \arctan x$.
- The interval is $[a, b] = [0, 1]$.
- The length of the interval is $b-a = 1 - 0 = 1$.
Substituting these into the formula, we need to calculate:
$$f_{avg}=\frac{1}{1-0}{\int }_0^1\arctan x\hspace{0.17em}dx={\int }_0^1\arctan x\hspace{0.17em}dx$$
f_{avg} = \frac{1}{1-0} \int_{0}^{1} \arctan x \, dx = \int_{0}^{1} \arctan x \, dx

The average value is the value of the definite integral $\int_{0}^{1} \arctan x \, dx$.


The integral $\int \arctan x \, dx$ is not a basic integral form. It can be solved using Integration by Parts, where we treat $\arctan x$ as one part and $dx$ (or $1 \cdot dx$) as the other part. The formula is $\int u \, dv = uv - \int v \, du$.


Let's choose $u$ and $dv$. According to the LIATE/LIPET guideline (Inverse Trig comes before Algebraic/Polynomial which is just 1 here), we choose:
Let:
$$u = \arctan x$$
$$dv = dx$$
Now, find $du$ by differentiating $u$, and find $v$ by integrating $dv$:
$$du = \frac{1}{1+x^2} dx$$
$$v = \int dx = x$$
Substitute these into the integration by parts formula:
$$\begin{align*}$$
\int \arctan x \, dx &= uv - \int v \, du \\
&= (\arctan x)(x) - \int x \left( \frac{1}{1+x^2} dx \right) \\
&= x \arctan x - \int \frac{x}{1+x^2} dx
\end{align*}
The remaining integral $\int \frac{x}{1+x^2} dx$ can be solved using a simple u-substitution.


Let $w = 1+x^2$. Then $dw = 2x \, dx$, which means $x \, dx = \frac{1}{2} dw$.
$$\begin{align*}$$
\int \frac{x}{1+x^2} dx &= \int \frac{1}{w} \left( \frac{1}{2} dw \right) \\
&= \frac{1}{2} \int \frac{1}{w} dw \\
&= \frac{1}{2} \ln|w| + C_1 \\
&= \frac{1}{2} \ln(1+x^2) + C_1
\end{align*}
(Absolute value is dropped since $1+x^2$ is always positive).


Substitute the result from Step 4a back into the integration by parts result:
$$\begin{align*}$$
\int \arctan x \, dx &= x \arctan x \\ &\quad - \frac{1}{2} \ln(1+x^2) + C
\end{align*}
The antiderivative is $F(x) = x \arctan x - \frac{1}{2} \ln(1+x^2)$.


Now we evaluate the definite integral using the antiderivative $F(x)$ and the limits $a=0, b=1$:
$$\begin{align*}$$
\int_{0}^{1} \arctan x \, dx &= \left[ x \arctan x - \frac{1}{2} \ln(1+x^2) \right]_{0}^{1} \\
&= F(1) - F(0)
\end{align*}
Evaluate $F(1)$:
$$\begin{align*} F(1) &= (1)\arctan(1) - \frac{1}{2} \ln(1+1^2) \\ &= 1 \cdot \frac{\pi}{4} - \frac{1}{2} \ln(2) \\ &= \frac{\pi}{4} - \frac{\ln 2}{2} \end{align*}$$
Evaluate $F(0)$:
$$\begin{align*} F(0) &= (0)\arctan(0) - \frac{1}{2} \ln(1+0^2) \\ &= 0 \cdot 0 - \frac{1}{2} \ln(1) \\ &= 0 - \frac{1}{2} \cdot 0 = 0 \end{align*}$$
Calculate the difference:
$$\begin{align*}$$
\int_{0}^{1} \arctan x \, dx &= F(1) - F(0) \\ &= \left( \frac{\pi}{4} - \frac{\ln 2}{2} \right) - 0 \\ &= \frac{\pi}{4} - \frac{\ln 2}{2}
\end{align*}


From Step 2, we found that $f_{avg} = \int_{0}^{1} \arctan x \, dx$.
Since the integral evaluates to $\frac{\pi}{4} - \frac{\ln 2}{2}$, this is the average value.


The average value of $f(x) = \arctan x$ over the interval $[0, 1]$ is $\frac{\pi}{4} - \frac{\ln 2}{2}$.
$$\boxed{{\displaystyle f_{avg}=\frac{\pi }{4}-\frac{\ln 2}{2}}}$$
\boxed{ f_{avg} = \frac{\pi}{4} - \frac{\ln 2}{2} }

1. Check for Convergence (Alternating Series Test):
This is an alternating series with $b_n = \frac{n}{n^2+1}$. We check the conditions for the Alternating Series Test (AST).

- Positivity: $b_n = \frac{n}{n^2+1} > 0$ for $n \ge 1$. Condition holds.

- Limit is zero: $\lim_{n\to\infty} b_n = \lim_{n\to\infty} \frac{n}{n^2+1} = \lim_{n\to\infty} \frac{1/n}{1+1/n^2} = \frac{0}{1+0} = 0$. Condition holds.

- Decreasing terms: Let $f(x) = \frac{x}{x^2+1}$. $f'(x) = \frac{(x^2+1)(1) - (x)(2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$. $f'(x) < 0$ when $1-x^2 < 0$, which means $x^2 > 1$, or $x > 1$. So, $b_n$ is decreasing for $n \ge 2$. Condition holds.

Since all conditions are met, the series $\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2+1}$ converges by the AST.

2. Check for Absolute Convergence:
We test the convergence of $\sum_{n=1}^{\infty} |a_n| = \sum_{n=1}^{\infty} \frac{n}{n^2+1}$.
Use the Limit Comparison Test (LCT). Compare $a_n = \frac{n}{n^2+1}$ with the harmonic series $b_n = \frac{1}{n}$.
$$\begin{align*} L &= \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n/(n^2+1)}{1/n} \\ &= \lim_{n\to\infty} \frac{n^2}{n^2+1} \\ &= \lim_{n\to\infty} \frac{1}{1+1/n^2} = \frac{1}{1+0} = 1 \end{align*}$$
Since $0 < L < \infty$ and the comparison series $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges (p-series with $p=1$), the series $\sum_{n=1}^{\infty} \frac{n}{n^2+1}$ also diverges by LCT.

3. Conclusion for (a):
The series converges (by AST), but it does not converge absolutely (by LCT with $1/n$).