1. ## Proof series diverges

$\displaystyle \sum_{n=1}^{\propto } \frac{3\sqrt{n}}{n}$

How to prove this series diverges?

Many thanks!

2. ## Re: Proof series diverges

Originally Posted by BabyMilo
$\displaystyle \sum_{n=1}^{\propto } \frac{3\sqrt{n}}{n}$

How to prove this series diverges?

Many thanks!
$\displaystyle \sum_{n=1}^\infty\frac{3\sqrt{n}}{n}=$$\displaystyle \sum_{n=1}^\infty\frac{3}{\sqrt{n}} \displaystyle \sum_{n=1}^\infty\frac{1}{\sqrt{n}} is a p-series with \displaystyle p=\frac{1}{2}. So, \displaystyle \sum_{n=1}^\infty\frac{1}{\sqrt{n}} diverges and consequently, \displaystyle \sum_{n=1}^\infty\frac{3}{\sqrt{n}} diverges. 3. ## Re: Proof series diverges Originally Posted by alexmahone \displaystyle \sum_{n=1}^\infty\frac{3\sqrt{n}}{n}=$$\displaystyle \sum_{n=1}^\infty\frac{3}{\sqrt{n}}$

$\displaystyle \sum_{n=1}^\infty\frac{1}{\sqrt{n}}$ is a p-series with $\displaystyle p=\frac{1}{2}$. So, $\displaystyle \sum_{n=1}^\infty\frac{1}{\sqrt{n}}$ diverges and consequently, $\displaystyle \sum_{n=1}^\infty\frac{3}{\sqrt{n}}$ diverges.
but when i tried to do a ratio test, the answer is smaller than 1 which suggests it converges, but we know this series diverges.
can you show me why?

many thanks.

4. ## Re: Proof series diverges

Originally Posted by BabyMilo
but when i tried to do a ratio test, the answer is smaller than 1 which suggests it converges, but we know this series diverges.
can you show me why?

many thanks.
Actually L = 1, which makes the ratio test inconclusive.

5. ## Re: Proof series diverges

Originally Posted by alexmahone
Actually L = 1, which makes the ratio test inconclusive.
is there another way of proofing?
other than the p-series and the ratio test?

6. ## Re: Proof series diverges

Originally Posted by BabyMilo
is there another way of proofing?
other than the p-series and the ratio test?
If you know that the harmonic series diverges, you can use the comparison test.

7. ## Re: Proof series diverges

A 'direct' proof that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges is very comfortable. If we consider the partial sum of the first k terms we note that...

$\displaystyle S_{k}=\sum_{n=1}^{k} \frac{1}{\sqrt{n}} \ge k\ \frac{1}{\sqrt{k}}= \sqrt{k}$ (1)

... so that $\displaystyle \lim_{k \rightarrow \infty} S_{k}= \infty$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. ## Re: Proof series diverges

$\displaystyle \frac{1}{\sqrt{n}}$ is non-negative monotone decreasing function.

and:

$\displaystyle \int\frac{1}{\sqrt{n}}dn=2\sqrt{n}+C$