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Let $u_n := \sum_{k=1}^{n} \frac{1}{n + \sqrt{k}}.$

Classical and easy: $u_n < 1$ and $\lim\limits_{n\to\infty} u_n = 1$.

But how to prove $u_n < u_{n+1}$ for every $n$? I have a rather long and unpleasant proof using standard calculus, but maybe some one has an elegant one?

If noone gives such a proof, I'll finally give my own.

1 Answers 1

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The "long and unpleasant" proof, presented as an undergraduate homework.

Prove for $n > 1$:

  1. $u_{n+1} - u_n = v_n - w_n$ with: $ v_n = \frac{1}{(n+1+\sqrt{n+1})} $ and $ w_n = \sum_{k=1}^{n} \frac{1}{(n+\sqrt{k})(n+1+\sqrt{k})} $

  2. $v_n > 1/(n+1) - 1/(n+1)^{3/2}$

  3. $v_n > 1/(n+1) - 1/n^{3/2}$

  4. $v_n > 1/n - 1/n^{3/2} - 1/n^2$

  5. $w_n < \int_0^n \frac{dk}{(n+\sqrt{k})^2}$

  6. $w_n < 2 \ln(1+1/\sqrt{n}) - 2/(1+\sqrt{n})$

  7. $w_n < 1/n - 4/(3n^{3/2}) + 2/n^2$

  8. $u_{n+1} - u_n > 1/(3n^{3/2}) - 3/n^2$

  9. $n \geq 81 \Rightarrow u_{n+1} > u_n$

Finish the proof.