- #1

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

However I couldn't think of a simple way to prove that.

Can anyone prove that this equation holds true?

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- B
- Thread starter Imaxx
- Start date

- #1

- 5

- 0

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

However I couldn't think of a simple way to prove that.

Can anyone prove that this equation holds true?

- #2

- 35

- 35

$$

\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.

- #3

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\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}[/tex] 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.

- #4

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This was a post in Calc & Beyond HW

- #5

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- please report homework questions in technical forums, instead of answering them
- do not provide full answers, that doesn't help the OP to understand their problem, even in technical forums
- do not open multiple threads on the same topic
- homework questions (anyway where they have been posted) require some efforts to be shown from the OP. We are not a solution automaton. Our goal is to teach, not to solve.

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