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Let $S_n$ represent its partial sum, and let $s$ represent its value. Prove that $s$ is finite and find and an $n$ so large that $S_n$ approximates s to 3 decimal places. $$\sum_{k=1}^{\infty}\left(\frac{k}{k+1}\right)^{k^2}$$

Solution we use root test i think. But how to calculate this? Thanks.

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marked as duplicate by Gerry Myerson, Ayman Hourieh, Git Gud, Daniel Rust, Davide Giraudo Jun 30 '13 at 10:35

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1 Answer 1

Hint: Let your series be $\sum_{1}^\infty a_k$.

Let $b_k=\left(\frac{k+1}{k}\right)^k$. Note that the sequence $(b_k)$ is increasing and has limit $e$. But we don't need to know that, it is enough to observe that $b_k \ge 2$.

We have $a_k =\left(\frac{1}{b_k}\right)^k$. Thus $a_k\le \frac{1}{2^k}$. So we can bound the truncation error by the tail of a familiar geometric series.

For just showing that the series converges, Comparison with $\sum \fac{1}{2^k}$ does the job. If you want to use one of Ratio Test or Root Test, both will work, but Root Test is simpler.

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