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So I have this inequality and I just can't figure out how to prove it:

Prove that ($\forall n\in \mathbb N)$ $$\sum_{k=1}^n \frac{1}{(k+1)\sqrt k}<2.$$ I've figured that for $n=1$ the inequality holds, since: $\frac{1}{2}<2$; so the statement is true for some $n \in \mathbb N$. Although I can't figure out how to prove the implication, that if the statement holds for some $n \in \mathbb N$, then it also holds for the number $n+1$.

I can't seem to make use of the induction hypothesis since the right-hand side is just a constant.

I'd be grateful for any suggestions.

edit: Solved, thanks for the answers.

Martin Sleziak
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David
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  • Compare it with a suitable integral? It's the first thing I would look to, at least. – Harald Hanche-Olsen Jan 08 '14 at 21:05
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    Find a sequence $c_n > 0$ such that $$\sum_{n=1}^N \frac{1}{(n+1)\sqrt{n}} < 2 - c_N.$$ – Daniel Fischer Jan 08 '14 at 21:05
  • Looks like $n$ in the series is a dummy index.. But you don't refer to it as such.. Can you clarify? – user76568 Jan 08 '14 at 21:07
  • I'm sorry, seems that I made a mistake, what I meant is that it holds for any $n \in \mathbb N$ from $k=1$ to $n$, thanks for noticing. I've changed it in the post. – David Jan 08 '14 at 21:16
  • @David : this problem is equivalent to showing $\sum_{k=1}^\infty \frac{1}{(k+1)\sqrt{k}} \leq 2$. – Stefan Smith Jan 09 '14 at 02:38
  • @stefan : Well, yes but then is that really easier? And if so, how should I proceed? – David Jan 09 '14 at 07:39
  • @daniel I was trying some but without success, I tried $\frac{1}{k^2}$ and also a few others, none of them worked. I assume it should be something of the form $\frac{1}{Q(k)}$ or maybe $\frac{constant}{Q(k)}$, where $Q(k)$ is a polynomial but I'm not sure. – David Jan 09 '14 at 07:54
  • @David : read Igor's answer. Note that the terms are decreasing. Compare the infinite series to the area under a curve (given by an improper integral). Draw pictures. – Stefan Smith Jan 09 '14 at 18:54
  • @StefanSmith The thing is I haven't learnt about that yet, so I don't know how to convert this problem into integrals. I am sure it is possible to solve it using induction or by direct comparison to a convergent series with sum < 2 I just can't figure out how. These kinds of problems will be on our exam but as I mentioned, we haven't covered integrals yet, so there must be a simpler way to solve it. – David Jan 09 '14 at 19:08

2 Answers2

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Note that

$$\frac1{\sqrt{k}}-\frac1{\sqrt{k+1}}=\frac1{\sqrt{k}\sqrt{k+1}(\sqrt{k}+\sqrt{k+1})}\ge\frac1{2(k+1)\sqrt{k}}.$$

Lutz Lehmann
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  • So the LHS is basically a telescoping series, which converges to 1 and after multiplying the inequality by 2 I get the desired result, right? Actually the "$\ge$" sign can be changed to a "$>$" sign since it holds for every k, to fit the original problem. I just wanted to ask about the equality on the left side, could you elaborate on that a bit? Thanks a lot, by the way. – David Jan 09 '14 at 21:36
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    Nveremind, I figured out: $\frac1{\sqrt{k}}-\frac1{\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt k}{\sqrt k \sqrt{k+1}}=\frac{(\sqrt{k+1}-\sqrt k)(\sqrt{k+1}+\sqrt k)}{(\sqrt k \sqrt{k+1})(\sqrt{k+1}+\sqrt k)}=\frac1{\sqrt{k}\sqrt{k+1}(\sqrt{k}+\sqrt{k+1})}$. – David Jan 09 '14 at 21:49
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Estimate the $\sum_{n=N}^\infty$ by the integral of the function (from $N$ to $\infty$). Notice that your function is decreasing and convex, so you have two-sided bounds.

Igor Rivin
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