I'm just beginning my journey with calculus, and this problem is giving me hard time.

$$\lim_{n \to \infty} \sum_{k=1}^n \frac{k}{{n^2+k}}$$

I calculated first three sums: $$\frac{1}{2}, \frac{16}{30}, \frac{117}{220}$$ From that I guessed that the limit is $$\frac{1}{2}$$. The hard part, though, is proving it.

I reckon that the squeeze theorem should be used here, but I can't seem to find anything useful. (The only sequences I was able to figure out give me limits of 1 and 0, so it's not something I can use)

Maybe you guys can give me some ideas or theorems that can be of use here? I know that this problem could be probably solved using some more advanced calculus techniques, but I'm not familiar with derivatives and integrals yet, and this problem is in the limits section of my problem book, so I'm not supposed to use anything more advanced.

  • $\begingroup$ Are you sure the $k$ in the denominator isn't actually $k^2$? $\endgroup$ – Nameless Jan 6 '13 at 16:24

For every $1\leqslant k\leqslant n$, $n^2\leqslant n^2+k\leqslant n^2+n$ hence the $n$th sum $S_n$ is such that $$ \sum_{k=1}^n\frac{k}{n^2+n}\leqslant S_n\leqslant\sum_{k=1}^n\frac{k}{n^2}. $$ Now, if it happens that the lower bound and the upper bound both converge to the same limit $\ell$, then $\lim\limits_{n\to\infty}S_n=\ell$.

Hence, what is left to do is to show that $\ell$ exists and to compute its value (or rather, to check that your guess that $\ell=\frac12$ is indeed correct--which it is).

  • 3
    $\begingroup$ Thank you so much for this helpful answer :) The solution always looks easy when you see it, but I really struggled until I read your response. Mathematics is so wonderful :) $\endgroup$ – TerryGoldenstein Jan 6 '13 at 16:37
  • $\begingroup$ Thanks for the appreciation. $\endgroup$ – Did Jan 6 '13 at 19:16

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