# Does $\sum_{k=1}^{\infty} \frac{1}{k\sqrt{\vphantom{} k+1}}$ converge?

Does the following series converge? $$\sum_{k=1}^{\infty} \frac{1}{k\sqrt{\vphantom{|} k+1}}$$

I tried using the ratio test and the comparison test but I wasn't able to solve this.

I think I should try manipulating the denominator to use comparison test but I can't figure out how?

• Do you know p-test? Sep 8, 2016 at 12:34
• Correct. You should manipulate the denominator. Sep 8, 2016 at 12:34
• @ThePortakal no, I'm not familiar with that Sep 8, 2016 at 12:36
• @112358 The p-test says $\sum \frac{1}{k^p}$ converge if $p>1$, and diverge if $p \leq 1$ Sep 8, 2016 at 12:37
• @snulty: Why is there a \vphantom there? Sep 9, 2016 at 12:59

Note that $\frac{1}{k\sqrt{k+1}} < \frac{1}{k \sqrt{k}} = k^{-1.5}$.

Hence, by the comparison test, $\displaystyle\sum_{k=1}^n\frac{1}{k\sqrt{k+1}} < \displaystyle\sum_{k=1}^n k^{-1.5} < \infty$ for all $n$. Hence $\displaystyle\sum_{k=1}^n\frac{1}{k\sqrt{k+1}}$ converges, and in fact it is somewhere close to $2.04288$ by Wolfram Alpha.

• Mathematica says that the sum is approximately $2.18400947\dots$
– robjohn
Sep 8, 2016 at 15:23
• I didn't take enough terms. Anyway, it's convergent. Sep 8, 2016 at 23:33
• perhaps a typo: $\frac1{k\sqrt{k}}\color{#C00000}{=}k^{-1.5}$
– robjohn
Sep 9, 2016 at 6:37
• @robjohn Thank you for pointing out the mistake. Sep 9, 2016 at 6:39
• $\sum_n n^{-p}$ converges for $p>1$ by the Cauchy Condensation Test. Sep 9, 2016 at 11:38

A slight variation of robjonh's fine answer through creative telescoping.
We may check in advance that for every $k\geq 1$ the inequality $$\frac{1}{k\sqrt{k+1}}\leq \frac{2}{\sqrt{k-\frac{1}{5}}}-\frac{2}{\sqrt{k+\frac{4}{5}}}\tag{1}$$ holds, hence it follows that: $$\sum_{k\geq 1}\frac{1}{k\sqrt{k+1}}\leq \frac{2}{\sqrt{1-\frac{1}{5}}}=\color{red}{\sqrt{5}}. \tag{2}$$

• This gets closer to the actual sum of $2.18400947\dots$ than $1+\sqrt2$, which I got. (+1) What's a good way to verify $(1)$?
– robjohn
Sep 8, 2016 at 14:13
• @robjohn: I "squared till death" and computed a discriminant, but there are probably slicker ways. Sep 8, 2016 at 14:18

Taking the terms in groups of $2^n$, $$\frac1{\sqrt2}<1,\\ \frac1{2\sqrt3}+\frac1{3\sqrt4}<\frac1{\sqrt2},\\ \frac1{4\sqrt5}+\frac1{5\sqrt6}+\frac1{6\sqrt7}+\frac2{7\sqrt8}<\frac1{\sqrt{2^2}},\\ \frac1{8\sqrt9}+\frac1{9\sqrt{10}}+\frac1{10\sqrt{11}}+\frac2{11\sqrt{12}}+\frac1{12\sqrt{13}}+\frac1{13\sqrt{14}}+\frac1{14\sqrt{15}}+\frac2{15\sqrt{16}}<\frac1{\sqrt{2^3}},\\\cdots$$ and the sum is bounded by a geometriec series of common factor $1/\sqrt2$.

The property will remain true replacing the square root by any positive power. (And by a similar lower bound you will show divergence for any non-positive power.)

For $k\ge1$, $\sqrt{k+1}\le\sqrt{2k}$. Therefore, \begin{align} \frac1{\sqrt{k}}-\frac1{\sqrt{k+1}} &=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}\sqrt{k+1}}\\ &=\frac1{\sqrt{k}\sqrt{k+1}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ &\ge\frac1{\left(1+\sqrt2\right)k\sqrt{k+1}} \end{align} Thus, \begin{align} \sum_{k=1}^\infty\frac1{k\sqrt{k+1}} &\le\left(1+\sqrt2\right)\sum_{k=1}^\infty\left(\frac1{\sqrt{k}}-\frac1{\sqrt{k+1}}\right)\\ &=1+\sqrt2 \end{align} So the series converges.

Using $8$ terms of the Euler-Maclaurin Sum Formula applied to $14$ terms of the Taylor Series for $\frac1{k\sqrt{k+1}}$ and comparing to $1000$ terms of the actual sum gives $$\sum_{k}\frac1{k\sqrt{k+1}}=2.18400947026785195289473415785294907$$

Using the fact that square root of x+1 > square root of (x), for x>0 and p test, it converges

• You think that $\sum\frac1{n\log n}$ converges?
– Did
Sep 9, 2016 at 12:44
• @TravisJ Yeah, a silent edit replaced ln by sqrt... Gotta love it. (Note: This also copies astonvilla's answer.)
– Did
Sep 9, 2016 at 13:42