another sum of series quesiton

[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 ?

[whozum]

Reevaluate a_2

[ILoveBaseball]

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

[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?

[ILoveBaseball]

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

[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

[Jameson]

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

[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.