# Proving the divergence of $\sum a_n$ with $a_{n}=\frac{1}{n}\left( 1+ \frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}} \right)$ [duplicate]

How can we prove formally that the series $\sum a_{n}$ diverges whose $n^{th}$ term has been provided below:

$$a_{n}=\frac{1}{n}\left( 1+ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}} \right),\qquad n\ge1.$$

The sequence converges to zero but I am not sure how book has proved that the series diverges?

## marked as duplicate by Martin R, Daniel Fischer sequences-and-series 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(); } ); }); }); Jan 3 '18 at 15:15

• As a sidenote: one can estimate $a_n$ using that it's basically a Riemann sum: $a_n = \frac{1}{n}\sum_{i=1}^n \frac{1}{\sqrt{i/n}}\cdot \frac{1}{\sqrt{n}} \sim \frac{1}{\sqrt{n}}\int_0^1 \frac{dx}{\sqrt{x}}$ – Winther Jan 3 '18 at 11:06
• Find a more general approach here:math.stackexchange.com/questions/2587694/… – user503348 Jan 3 '18 at 11:31
• Stolz-Ces$\mathrm{\grave{a}}$ro Theorem: $\displaystyle\lim_{n \to \infty}a_{n} = \lim_{n \to \infty}{1 \over \,\sqrt{\, n + 1\, }\,} = \color{red}{0}$. – Felix Marin Apr 19 '18 at 5:11

We have $$a_n\ge\frac1n,\quad n=1,2,3,\cdots,$$ thus the given series $\sum a_n$ diverges by comparison test with the harmonic series.

A bit of detail :

$a_n =$

$(1/n)(1+1/√2+1√3 +...1/√n)$

$\gt (1/n)( n/√n) =1/√n \gt 1/n.$

$S_n := \sum_{k=1}^{n} 1/k$ diverges.

• You can simply say that $\left( 1+ \frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}} \right)\ge1$ giving $\frac1n\left( 1+ \frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}} \right)\ge \frac1n.$ – Olivier Oloa Jan 3 '18 at 11:32
• Olivier. Very nice . – Peter Szilas Jan 3 '18 at 14:16