# Proof: $2\sqrt{m}-2 < \sum\limits_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1$

I know that problems similar to this one, involving either one of the two bounds, have been posted before, but I would like just a hint in the last part of the proof involving the upper bound, with which I have some troubles with being ''convincing'' enough.

Prove that $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$ if $n\geq 1$. Then use this to prove that

$$2\sqrt{m}-2 < \sum_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1 \qquad \text{if }\ m \geq 2$$

With some help I received here I got the following proof for the first set of inequalities:

Proof (direct):

Since

$$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{1}{2\sqrt{n}}$$

then we have

$$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}\qquad (1)$$

Likewise, since

$$\frac{1}{2\sqrt{n}} = \frac{1}{\sqrt{n} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n-1}} = \sqrt{n} - \sqrt{n-1}$$

then

$$\frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})\qquad (2)$$

as asserted.

For the second part I tried to prove the lower bound firt:

Proof (direct): From $(1)$ we have

$$\sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}}$$

By summing both sides we get

$$\sum_{n=1}^{m}(\sqrt{n+1} - \sqrt{n}) < \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}$$

And after applying the telescoping property on the LHS, this becomes

$$\sqrt{m+1} - 1 < \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}$$

But since $\sqrt{m} - 1 < \sqrt{m+1} - 1$, by transitivity we have

\begin{align*}\sqrt{m} - 1 &< \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}\\ 2\sqrt{m} - 2 &< \sum_{n=1}^{m}\frac{1}{\sqrt{n}}\end{align*}

Then the upper bound:

Proof (direct): From $(2)$ we have

$$\frac{1}{2\sqrt{n}}< \sqrt{n} - \sqrt{n-1}$$

By taking the sum of both sides we get

$$\sum_{n=1}^{m}\frac{1}{2\sqrt{n}}< \sum_{n=1}^{m}(\sqrt{n} - \sqrt{n-1})$$

If we apply the telescoping property on the RHS, the inequality becomes

\begin{align*}\sum_{n=1}^{m}\frac{1}{2\sqrt{n}} &< \sqrt{n}\\ \sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 2\sqrt{n}\end{align*}

Here I'm not sure how to get the $-1$ on the RHS that I'm lacking. I could have started the summation at $n=2$ I guess, but don't think it's correct and it doesn't seem to be a convincing way to complete the proof.

Well you tagged induction so why not use it:

$$\sum_{n=1}^{m+1} \frac{1}{\sqrt{n}} \lt 2\sqrt{m}-1 + \frac{1}{\sqrt{m+1}} \lt 2\sqrt{m}-1 + 2(\sqrt{m+1}-\sqrt{m}) = 2\sqrt{m+1}-1$$

• @Jazz yes, actually starting the summation at $n=2$ works like a charm. Why did you think it was incorrect? – benji Feb 9 '15 at 21:07

For the lower bound, proof (direct): from $$(1)$$:

$$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}$$

By summing both sides:

$$2\sum_{n=1}^{m}(\sqrt{n+1} - \sqrt{n}) < \sum_{n=1}^{m}\frac{1}{\sqrt{n}}$$

After applying the telescoping property on the LHS, the inequality becomes

$$2\sqrt{m+1} - 2 < \sum_{n=1}^{m}\frac{1}{\sqrt{n}}$$

For the upper bound, proof (direct): from $$(2)$$:

$$\frac{1}{\sqrt{n}}< 2(\sqrt{n} - \sqrt{n-1})$$

By taking the sum of both sides from n=2:

$$\sum_{n=2}^{m}\frac{1}{\sqrt{n}}< \sum_{n=2}^{m}2(\sqrt{n} - \sqrt{n-1})$$

$$1+\sum_{n=2}^{m}\frac{1}{\sqrt{n}}< 1+2(\sum_{n=2}^{m}(\sqrt{n} - \sqrt{n-1}))$$

$$\sum_{n=1}^{m}\frac{1}{\sqrt{n}}< 1+2(\sum_{n=2}^{m}(\sqrt{n} - \sqrt{n-1}))$$

Apply the telescoping property on the RHS, the inequality becomes

\begin{align*}\sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 1+2(\sqrt{m}-1)\\ \sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 2\sqrt{m}-1\end{align*}

So $$2\sqrt{m+1} - 2 < \sum_{n=1}^{m}\frac{1}{\sqrt{n}} < 2\sqrt{m} - 1$$

This can be relaxed to: