I am having a hard time proving if this sum converges or diverges:
$$\sum_{x=1}^{\infty} \frac{1}{\sqrt{x}\sqrt{x+1}}$$
I tried proving it by the ratio test but $q = 1$.
I couldn’t proceed further and would like some help.
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is $k=x$? $ $$ $$ $$ $ – Calvin Khor Apr 28 '22 at 11:56
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$\sqrt {k+1} \leq \sqrt {2k}$ – Kavi Rama Murthy Apr 28 '22 at 12:00
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@CalvinKhor sure, fixed – jophny Apr 28 '22 at 12:05
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$\sqrt{x(x+1)}<x+1$. – Gary Apr 28 '22 at 12:08
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Dunno if it helps, but $\sqrt x<\sqrt {x+1}$ because the square root is increasing, and therefore
$$\frac1{\sqrt x\sqrt{x+1}}>\frac1{\sqrt {x+1}\sqrt {x+1}}=\frac1{x+1}$$
because $x\mapsto 1/x$ is decreasing for positive $x$. Thus you are left with the harmonic series (minus 1).
emacs drives me nuts
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So you can show that $$\sum_{x=1}^{\infty} \frac{1}{x+1} < \sum_{x=1}^{\infty} \frac{1}{\sqrt{x}\sqrt{x+1}} < \sum_{x=1}^{\infty} \frac{1}{x} $$ – kronenbouh Apr 28 '22 at 12:31
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@kronenbouh Yes, but the left-hand inequality is enough because the harmonic series grows beyond all positive bounds. – emacs drives me nuts Apr 28 '22 at 12:37
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@kronenbouh ...and the $<$ turns into a $\leqslant$ in the limit. – emacs drives me nuts Apr 28 '22 at 12:43
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