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I still haven't gotten the hang of how to solve these problems, but when I first saw this one I thought partial fraction or limit. So I went with taking the limit but the solution manual shows them using the integral test.

Was I wrong to just take the limit?

$$\sum_{n=1}^{\infty}\frac{1}{n^2+4}$$

Next:

$$\lim_{n\to\infty}\frac{1}{n^2+4}=0$$

So converges by the test for divergence?

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  • $\begingroup$ The test for divergence can only show you that a series diverges if the limit is not zero. If the limit of the size of the terms is zero, then the divergence test is inconclusive. That's why they call it the divergence test. $\endgroup$ – G Tony Jacobs Nov 23 '13 at 4:07
  • $\begingroup$ When $\displaystyle{\large n \gg 2, {1 \over n^{2} + 4} \sim {1 \over n^{2}}}$ which is the Basel Problem. It converges. $\endgroup$ – Felix Marin Nov 23 '13 at 5:50
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We only have the following statement to be true: $$\text{If $\sum_{n=1}^{\infty} a_n$ converges, then $a_n \to 0$.}$$ The converse of the above statement is not true, i.e., $$\text{if $a_n \to 0$, then $\displaystyle \sum_{n=1}^{\infty} a_n$ converges is an incorrect statement.}$$

For instance, $\displaystyle \sum_{n=1}^{\infty} \dfrac1n$ diverges, even though $\dfrac1n \to 0$.

To prove your statement, note that $\dfrac1{n^2+4} < \dfrac1{n^2}$ and make use of the fact that $\displaystyle \sum_{n=1}^{\infty} \dfrac1{n^2}$ converges to conclude that $\displaystyle \sum_{n=1}^{\infty}\dfrac1{n^2+4}$ converges.

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  • $\begingroup$ Awesome! I see exactly what you mean, I assumed too much was true when in fact its only true that the sequence approaches 0, not the series? $\endgroup$ – hax0r_n_code Nov 23 '13 at 4:06
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Just expounding on $\frac{1}{n^2 + 4} < \frac{1}{n^2}$ in the above answer.

This holds using the Comparison Test which states that if $\sum a_n$ and $\sum b_n$ are such that $0 \le a_n \le b_n$, if $\sum b_n$ converges, then $\sum a_n$ converges.

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  • 1
    $\begingroup$ So I could use this test or the integral test then? $\endgroup$ – hax0r_n_code Nov 23 '13 at 4:11
  • $\begingroup$ To use the Integral Test, $f(x)$ which in this case would be $\frac{1}{x^2 +4}$ has to be positive ($f(x) \ge 0$), non-increasing on $(N, \infty \rbrace$. Then you'd find the limit of the integral of $f(x)$. If the limit $\to$ a constant then $\sum a_n$ converges. So it'd work for this problem. $\endgroup$ – Zhoe Nov 23 '13 at 4:23
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Another way, that needs fewer theorems:

$n^2+4 > n(n-1)$ so $\frac1{n^2+4} < \frac1{n(n-1)} = \frac1{n-1}-\frac1{n} $.

Therefore, for any $m > 0$ $\sum_{n=2}^m \frac1{n^2+4} < \sum_{n=2}^m (\frac1{n-1}-\frac1{n}) = 1-\frac1{m} < 1 $.

Therefore $\sum_{n=2}^m \frac1{n^2+4}$ converges as $m \to \infty$.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} \sum_{n = 1}^{\infty}{1 \over n^{2} + 4} &= \sum_{n = 0}^{\infty}{1 \over \pars{n + 1 + 2\ic}\pars{n + 1 - 2\ic}} = {\Psi\pars{1 + 2\ic} - \Psi\pars{1 - 2\ic} \over \pars{1 + 2\ic} - \pars{1 - 2\ic}} = {1 \over 2}\,\Im\Psi\pars{1 + 2\ic} \\[3mm]&= {1 \over 2}\bracks{-\,{1 \over 4} + {1 \over 2}\,\pi\coth\pars{2\pi}} \end{align} $$ \color{#0000ff}{\large\sum_{n = 1}^{\infty}{1 \over n^{2} + 4}} = \color{#0000ff}{\large{1 \over 4}\bracks{\pi\coth\pars{2\pi} - {1 \over 2}}} \approx 0.6604 $$

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