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I want ask a question about a sum. The exercise is as follows:

Prove the following inequality for every $n \geq 1$:

$$\sum\limits_{k=1}^n \frac{1}{k^2+3k+1} \leq \frac{13}{20} .$$

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    Related post: http://math.stackexchange.com/q/84899/. (By the way, @Maria, if you posted the other question through a different account, then you could merge the two accounts by flagging the moderators. If that was posted by someone else, please ignore my comment. ;))2011-11-24
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    ${1 \over k^2 + 3k + 1} < {1 \over k^2 + 3k}$. Now use partial fractions on ${1 \over k^2 + 3k}$.2011-11-24
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    Just to nitpick, that $i$ should really be $k$ :)2011-11-24
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    Using Zarrax' comment, and writing down the resulting telescoping series, you'll see the partial sums are bounded by 11/18.2011-11-24
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    A quick way is to notice that $$\frac{1}{k^2+3k+1}\leq \frac{1}{k^2+2k+1}=\frac{1}{(k+1)^2}.$$ Then extending the series to infinity, this is $$\sum_{k=1}^\infty \frac{1}{(k+1)^2}=\zeta(2)-1=\frac{\pi^2}{6}-1\approx 0.644934\leq 0.65 =\frac{13}{20}.$$2011-11-24
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    As an aside, $~\displaystyle\sum_{k=1}^\infty\frac1{k^2+3k+1} ~=~ \frac\pi{\sqrt5}~\tan\bigg(\sqrt5~\frac\pi2\bigg)$.2015-03-11

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