2. Using the Integral Test, determine whether the following series converge. a) $$ \sum_{k=1}^{\infty} \frac{k}{k^{2}+1} $$ b) $$ \sum_{k=3}^{\infty} \frac{1}{\sqrt{2 k-5}} $$ c) $$ \sum_{k=0}^{\infty} \frac{1}{k^{2}+4} $$

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a)
We examine the series
$$
\sum_{k=1}^{\infty} \frac{k}{k^{2}+1}.
$$
Define
$$
f(x)=\frac{x}{x^{2}+1}.
$$
For $$x\ge 1$$, $$f(x)$$ is positive, continuous, and (for large $$x$$) eventually decreasing. To apply the Integral Test, we consider
$$
\int_1^\infty \frac{x}{x^2+1}\,dx.
$$

Let
$$
u=x^2+1,\quad \text{so that} \quad du=2x\,dx \quad \Longrightarrow \quad x\,dx=\frac{1}{2}\,du.
$$
Changing the limits: when $$x=1$$, $$u=1^2+1=2$$; when $$x\to\infty$$, $$u\to\infty$$. The integral becomes
$$
\int_1^\infty \frac{x}{x^2+1}\,dx=\frac{1}{2}\int_{2}^{\infty}\frac{1}{u}\,du.
$$
Evaluate the integral:
$$
\frac{1}{2}\int_{2}^{\infty}\frac{1}{u}\,du=\frac{1}{2}\left[\ln|u|\right]_{2}^{\infty}=\frac{1}{2}\lim_{t\to\infty}\bigl(\ln t - \ln 2\bigr).
$$
Since $$\ln t$$ diverges as $$t\to\infty$$, the integral diverges. By the Integral Test, the series diverges.

---

b)
Consider the series
$$
\sum_{k=3}^{\infty} \frac{1}{\sqrt{2k-5}}.
$$
Define
$$
f(x)=\frac{1}{\sqrt{2x-5}}.
$$
For $$x\ge 3$$, $$f(x)$$ is positive, continuous, and decreasing. Apply the Integral Test by evaluating
$$
\int_3^\infty \frac{1}{\sqrt{2x-5}}\,dx.
$$

Make the substitution:
$$
u=2x-5\quad \Longrightarrow \quad du=2\,dx\quad\Longrightarrow\quad dx=\frac{du}{2}.
$$
When $$x=3$$, $$u=2\cdot 3-5=1$$; when $$x\to\infty$$, $$u\to\infty$$. The integral becomes
$$
\int_3^\infty \frac{1}{\sqrt{2x-5}}\,dx=\frac{1}{2}\int_1^\infty \frac{1}{\sqrt{u}}\,du.
$$
Write the integrand as $$u^{-1/2}$$:
$$
\frac{1}{2}\int_1^\infty u^{-1/2}\,du.
$$
Integrate:
$$
\int u^{-1/2}\,du = 2u^{1/2}.
$$
Thus,
$$
\frac{1}{2}\cdot \left[2u^{1/2}\right]_1^\infty = \left[u^{1/2}\right]_1^\infty = \lim_{u\to\infty}\sqrt{u}-\sqrt{1}.
$$
Since $$\lim_{u\to\infty}\sqrt{u}=\infty$$, the integral diverges. Hence, by the Integral Test, the series diverges.

---

c)
Consider the series
$$
\sum_{k=0}^{\infty} \frac{1}{k^{2}+4}.
$$
Define
$$
f(x)=\frac{1}{x^2+4}
$$
for $$x\ge 0$$. For $$x\ge 0$$, $$f(x)$$ is positive, continuous, and decreasing. Apply the Integral Test by evaluating
$$
\int_0^\infty \frac{1}{x^2+4}\,dx.
$$

Recall the integral formula:
$$
\int \frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right)+C.
$$
Here, $$a^2=4$$ so $$a=2$$. Thus,
$$
\int_0^\infty \frac{1}{x^2+4}\,dx = \frac{1}{2}\arctan\left(\frac{x}{2}\right)\Bigg|_0^\infty.
$$
Evaluating the limits:
$$
\lim_{x\to\infty}\arctan\left(\frac{x}{2}\right)=\frac{\pi}{2}, \quad \arctan(0)=0.
$$
Therefore,
$$
\int_0^\infty \frac{1}{x^2+4}\,dx = \frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}.
$$
Since the integral converges, by the Integral Test the series converges.

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a) Diverges  
b) Diverges  
c) Converges
a) The series diverges. b) The series diverges. c) The series converges.

Using the Integral Test, determine whether the following series converge.
a)
b)
c)

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