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Consider the following sum:

$$\sum_{k=1}^\infty\prod_{j=1}^k\frac{1}{\sqrt{j+1}-\sqrt{j}+1}$$

How could I check whether this sum converge or diverge? Root and ratio tests are inconclusive...

JohnWO
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3 Answers3

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You have the equivalence (for $j\gg 1$) :

$$\sqrt{j+1}-\sqrt{j}+1 \sim 1+\sqrt{j}\left(\sqrt{1+\frac 1j}-1\right)\sim 1+\frac 1{2\sqrt{j}}-\frac 1{8j\sqrt{j}}$$ So that $$\log\left(\sqrt{j+1}-\sqrt{j}+1\right)\sim \frac 1{2\sqrt{j}}-\frac 1{8j}$$ and $$\prod_{j=1}^k\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\sim \exp\left[C-\sum_{j=1}^k \frac 1{2\sqrt{j}}-\frac 1{8j}\right]\sim D\;e^{-\sqrt{k}+(\ln k)/8}\sim D\;\sqrt[8]{k}\;e^{-\sqrt{k}}$$ I'll let you conclude.

Raymond Manzoni
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  • (+1) I was just about to transfer a very similar argument from my pad (of paper). – robjohn Aug 07 '13 at 19:48
  • Thanks @robjohn ! I am just annoyed by the $-\frac 1{8j}$ next term in the first line that gives a $\ln(k)/8$ at the end. Technically it is right since $\ln(k)$ is smaller than $\sqrt{k}$ but the $D$ constant is weaker. What would you have written at the end ? (you are usually more rigorous than I :-)) – Raymond Manzoni Aug 07 '13 at 19:50
  • Sorry for not responding before you took care of the $\frac18\log(k)$. I was trying to come up with a different approach. It came out more similar than I was hoping. If it is too similar, I will delete it. I think you could use that $1-x\le e^{-x}$ and not worry about the further terms. – robjohn Aug 07 '13 at 20:40
  • Thanks for your answer @robjohn (+1 of course and it is rather different so no problem)! I thought that you would prefer working with inequalities (I'll admit that it is probably more adapted to prove convergence). – Raymond Manzoni Aug 07 '13 at 20:45
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$$ \begin{align} \frac1{\sqrt{j+1}-\sqrt{j}+1} &=\frac{\sqrt{j+1}+\sqrt{j}}{1+\sqrt{j+1}+\sqrt{j}}\\ &=1-\frac1{1+\sqrt{j+1}+\sqrt{j}}\\ &\le1-\frac1{3\sqrt{j+1}} \end{align} $$ Since $1-x\le e^{-x}$ for all $x$, $$ \begin{align} \prod_{j=1}^k\left(1-\frac1{3\sqrt{j+1}}\right) &\le\exp\left(-\sum_{j=1}^k\frac1{3\sqrt{j+1}}\right)\\ &\le\exp\left(-\frac13\int_2^{k+2}\frac{\mathrm{d}x}{\sqrt{x}}\right)\\ &\le\exp\left(-\frac23\left(\sqrt{k+2}-\sqrt2\right)\right)\\ &\le e^{\frac23\sqrt2}e^{-\frac23\sqrt{k+2}} \end{align} $$ Noting the critical points at $u=0$ and $u=3$, we get that $$ u^3e^{-u}\le27e^{-3} $$ Therefore, $$ e^{\frac23\sqrt2}e^{-\frac23\sqrt{k+2}}\le\frac{e^{\frac23\sqrt2}\,27e^{-3}}{\left(\frac23\sqrt{k+2}\right)^3} $$ Now use the $p$-test with $p=\frac32$.

robjohn
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A different solution, which is not mine, is the following:

Let $\alpha_{k}=\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\,,\; k\in\mathbb{N}$, then \begin{align*} 1-\frac{\alpha_{k+1}}{\alpha_k}&=1-\frac{\prod_{j=1}^{k+1}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}}{\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}}\\ &=1-\frac{1}{\sqrt{k+2}-\sqrt{k+1}+1}\\ &=\frac{\sqrt{k+2}-\sqrt{k+1}}{\sqrt{k+2}-\sqrt{k+1}+1}\\ &=\frac{k+2-k-1}{\big(\sqrt{k+2}-\sqrt{k+1}+1\big)\big(\sqrt{k+2}+\sqrt{k+1}\big)}\\ &=\frac{1}{\sqrt{k+2}+\sqrt{k+1}+1}\,. \end{align*} Because \begin{align*} \mathop{\lim}\limits_{k\to+\infty}\frac{k}{\sqrt{k+2}+\sqrt{k+1}+1}&=\mathop{\lim}\limits_{k\to+\infty}\frac{\sqrt{k}}{\sqrt{1+\frac{2}{k}}+\sqrt{1+\frac{1}{k}}+\frac{1}{\sqrt{k}}}=+\infty\,, \end{align*} by Raabe's criterion, we have that the series $\sum_{{k}=1}^{\infty}\big(\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\big)$ converges.

Grigorios Kostakos
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