Is the series $\sum_{k=1}^{\infty}\frac{\arctan(2k)}{1+4k^2}$ convergent?

[karush]

$\tiny{206.f3a.}$
$\textsf{Use the divergence Test to detemine whether the series is divergent}$
\begin{align}
\displaystyle
&\sum_{k=1}^{\infty}\frac{\arctan(2k)}{1+4k^2}\\
\textit{take limit}\\
=&\lim_{{k}\to{\infty}}\frac{\arctan(2k)}{1+4k^2}\\
\\
=&\frac{\arctan(\infty)}{\infty} =\frac{\pi/2}{\infty}=0\\
\therefore inconclusive
\end{align}
$\textsf{Use the Integral Test to detemine whether the series is divergent. positive and continuous terms}$
\begin{align}
\displaystyle
f(k)&=\frac{\arctan(2k)}{1+4k^2}\\
f'(k)&=\frac{2-\arctan(2k)8k}{(1+4k)^2}\\
\textit{as } {{k}\to{\infty}} \, f(k) \textit{ decreases}\\
\int_{1}^{\infty} \frac{\arctan(2k)}{1+4k^2}\,dk&
=\lim_{{k}\to{\infty}}\int_{1}^{b} \frac{\arctan(2k)}{1+4k^2}\,dk\\
&u=\arctan(2k) \therefore du=\frac{2}{1+4k^2}\\
\frac{1}{2}\int u \, du&=\frac{u^2}{4}\\
&=\lim_{{b}\to{\infty}}\frac{1}{4}
\left[(\arctan(2b))^2)-((\arctan(2))^2 )\right] \\
&=\frac{1}{4}\left[\left(\frac{\pi}{2}\right)^2
-(\arctan(2))^2\right] \\
\textit{finite values }\\
&\therefore
\int_{1}^{\infty} \frac{\arctan(2k)}{1+4k^2}\,dk
\textit{ converges }\\
&\therefore
\sum_{k=1}^{\infty}\frac{\arctan(2k)}{1+4k^2}
\textit{ converges }
\end{align}
$\textit{ just seeing where errors are and sugestions? }$

 

[MarkFL]

I would use a limit comparison test. We know:

$$\sum_{k=1}^{\infty}\frac{1}{4k^2}=\frac{\pi^2}{24}$$

Thus we check:

$$\lim_{k\to\infty}\left(\frac{\arctan(2k)}{4k^2+1}\cdot\frac{4k^2}{1}\right)=\frac{\pi}{2}$$

Since the limit is a finite number, we know the original series converges.

 

[karush]

I would use a limit comparison test. We know:

$$\sum_{k=1}^{\infty}\frac{1}{4k^2}=\frac{\pi^2}{24}$$

where does
$\displaystyle

\frac{1}{4k^2}$
come from?

 

[MarkFL]

where does
$\displaystyle

\frac{1}{4k^2}$
come from?

It is the $k$th term of a known convergent series. When we take the $k$th term of the given series and divide by this one, we get an expression for whom the limit at infinity goes to a finite value, thereby showing the given series converges.

 

[karush]

really appreciate the help
going to spend the rest of the month
doing d/c problems