How to do comparison test for $\sum \frac{1}{\sqrt{n}}$ $sum 1 / (sqrt(n))$ n=1 to $n = \infty$ $n=oo$ ?

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Answer 1 · 2015-05-15T22:29:50

The series diverges.

Let's begin with an inequality :

$\sqrt{n} \leq n,$ $sqrtn <= n,$ , with $n \geq 0$ $n >=0$

$\frac{1}{\sqrt{n}} \geq \frac{1}{n}$ $1/sqrtn>= 1/n$ , with $n \geq 1$ $n>=1$

$\infty \sum n = 1 \frac{1}{\sqrt{n}} \geq \infty \sum n = 1 \frac{1}{n}$ $sum_(n=1)^oo1/sqrtn>=sum_(n=1)^oo1/n$

$\infty \sum n = 1 \frac{1}{n}$ $sum_(n=1)^oo1/n$ is the harmonic series and it diverges.

Therefore, by comparison test, $\infty \sum n = 1 \frac{1}{\sqrt{n}}$ $sum_(n=1)^oo1/sqrtn$ diverges.

How to do comparison test for ∑1√nsum 1 / (sqrt(n)) n=1 to n=∞n=oo?