# Another sum of series quesiton

1. Apr 24, 2005

### ILoveBaseball

Consider the series

$$\sum_{n=1}^\infty \frac{n}{\sqrt{5n^2+5}}$$

Value ______

$$a_1 = .316227766, a_2 = 2/5, a_3 = .4242640687, a_4 = .4338609156$$

there doesnt seem to be any common ratio, so that means that this isnt a geometric series right?

well i think i can simplify the equation to:

$$\frac{1}{\sqrt{5}}\sum_{n=1}^\infty \frac{n}{\sqrt{(n^2+1)}}$$

hmm, that's as far as i got, can someone help me ?

Last edited: Apr 24, 2005
2. Apr 24, 2005

### whozum

Reevaluate $a_2$

3. Apr 24, 2005

### ILoveBaseball

i made the changes, but i still dont see a common ratio

4. Apr 24, 2005

### Hippo

$$\frac{1}{\sqrt{5}}\sum_{n=1}^\infty \frac{n}{\sqrt{(n^2+1)}}$$

That thing is only going to have a value if it's convergent, correct? I don't really remember much about this stuff.

If a series converges to a finite value, its sequence a must converge to zero.

Contrapositively, if the sequence a does not converge to zero, the series does not converge to a finite value.

Here, the sequence a is
$$\frac{n}{\sqrt{(n^2+1)}}$$,
which converges to unity, not zero, as n approaches infinity.
This implies that the series in question is divergent.

Right?

5. Apr 24, 2005

### ILoveBaseball

it converges for sure, it was the first question asked.

6. Apr 24, 2005

### arildno

It does not converge ILoveBaseball; you must have mistyped.

If $$a_{n}=\frac{n}{\sqrt{5n^{2}+5}}$$
then,
$$\lim_{n\to\infty}a_{n}=\frac{1}{\sqrt{5}}>0$$
But a necessary requirement for convergence of the series $$\sum_{n=1}^{\infty}a_{n}$$ is that we have $$\lim_{n\to\infty}a_{n}=0$$

7. Apr 24, 2005

### Jameson

What reason do you have to think it does converge? Prove it mathematically.

8. Apr 24, 2005

### ILoveBaseball

sorry, you guys are right. for some reason there was a glitch, even if i selected converge, i got 50% of the problem correct(which means i got the first question right). but if i selected diverge, i got 100% of the problem(two problems = 100%) correct.