How to prove this infinite series?

[Imaxx]

While transforming the equation of the Basel problem, the following infinite series was obtained.

$$\sum_{n=1}^{\infty} \frac{n^2+3n+1}{n^4+2n^3+n^2}=2$$

However I couldn't think of a simple way to prove that.
Can anyone prove that this equation holds true?

 

[William Crawford]

Your series can be put on the form of a telescope series and thereby summed up.
$$
\begin{align*}
\sum_{n=1}^\infty\frac{n^2+3n+1}{n^4+2n^3+n^2} &= \sum_{n=1}^\infty\frac{n^2+3n+1}{n^2(n+1)^2} \\
&= \lim_{N\rightarrow\infty}\sum_{n=1}^N\bigg\lbrace\frac{2n^2+3n}{(n+1)^2} - \frac{2(n-1)^2+3(n-1)}{n^2}\bigg\rbrace \\
&= \lim_{N\rightarrow\infty}\frac{2N^2+3N}{(N+1)^2} \\
&= 2,
\end{align*}
$$
where the last equality follows by successive use of l'Hopital's rule.

 

[pasmith]

You don't need l'Hopital's rule here; just observe $$\frac{2N^2 + 3N}{(N + 1)^2} = \frac{N^2(2 + \frac 3N)}{N^2(1 + \frac1N)^2} = \frac{2 + \frac 3N}{(1 + \frac1N)^2}$$ and the result follows from the proposition that the limit of a ratio is the ratio of the limits provided the limit of the denominator is not zero.

 

[benorin]

This was a post in Calc & Beyond HW

 

[fresh_42]

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