# How do I evaluate $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}$? [duplicate]

How do i evaluate this limit : $$\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+............+\frac{1}{\sqrt{n}}\right)$$ ?

Thank you for any help .

## marked as duplicate by quid♦, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 25 '17 at 0:43

• may you mean \frac {1}{(sqrt(n)} in LHS in the title ? – zeraoulia rafik Jul 4 '15 at 14:46
• It is bounded above by 2 so it certainly has a limit. – Paul Jul 4 '15 at 14:55
• also should you have $\frac{1}{\sqrt{k}}$ – Chinny84 Jul 4 '15 at 16:16

Let $a_n = 1 + \dfrac {1}{\sqrt{2}} + \cdots + \dfrac {1}{\sqrt{n}}$ and $b_n = \sqrt{n}$ for all $n\geq 1$. We have $$\lim_{n\to \infty} \dfrac {a_{n+1} - a_n} {b_{n+1} - b_n} = \lim_{n\to \infty} \dfrac {1} {\sqrt{n+1}(\sqrt{n+1} - \sqrt{n})} = \lim_{n\to \infty} \dfrac {\sqrt{n+1} + \sqrt{n}} {\sqrt{n+1}} = 2,$$ so by the Stolz-Cesàro theorem we get $\lim_{n\to \infty} \dfrac {a_n} {b_n} = 2$
We bound $s_n=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}$ by integrals. We get $$\int_1^n \frac{dt}{\sqrt{t}}\le s_n\le 1+\int_1^{n-1} \frac{dt}{\sqrt{t}}.$$ Ealuate the integrals, divide by $\sqrt{n}$, and Squeeze.