# Thread: finding values for convergence

1. ## finding values for convergence

I'd like to find for which values of x does
(1) sum n=1 to infinity (x^n)/n converge?
(2) sum n=1 to infinity (x^n)/(n^2) converge?
(3) sum n=1 to infinity (n^x)/(n^2 +1) converge?

Do i all do the ratio test and set the limit of it less than 1??? or is there different rules to consider for the each cases???

for instance for the (1) I took the ratio test and then I set the limit of it less than 1
lim n->oo (x^n+1)/(n+1) <1 which then becomes (x^n+1) has to be less than n+1
then x must be less than (n+1)^-(n+1) is this the right way of solving the problem??? help !!

2. Be careful when you apply the ration test. For example, in the first question, put $a_n(x) =\frac{x^n}n$. We have $\left|\frac{a_{n+1}(x)}{a_n(x)}\right|=\frac{|x|n} {n+1}$. Now you have to take the limit $n\to\infty$ (look at the proof if it's not clear).
The ratio test works for the second question, for the third find the $x$ such that the limit when $n\to\infty$ is $0$.

3. Originally Posted by girdav
for the third find the $x$ such that the limit when $n\to\infty$ is $0$.
Be careful indeed.
Notice that if $x=1$ then $\frac{n^1}{n^2+1}\to 0.$
But $\sum\limits_{n = 1}^\infty {\frac{n}{{n^2 + 1}}} \geqslant \sum\limits_{n = 1}^\infty {\frac{1}{{2n}}}$

4. Originally Posted by Plato
Be careful indeed.
Notice that if $x=1$ then $\frac{n^1}{n^2+1}\to 0.$
But $\sum\limits_{n = 1}^\infty {\frac{n}{{n^2 + 1}}} \geqslant \sum\limits_{n = 1}^\infty {\frac{1}{{2n}}}$
I know. I gave this "hint" because it's not easy to see the convergence without any computation (but the problem is that in this case the might be a lot of $x$ such that $a_n(x)\rightarrow 0$ and the series is not convergent). I realize it's not the shortest way to solve it. If $x\leq 0$ the series is convergent, and if $x>0$ the series has the same behavior as $\sum_{n=1}^{+\infty}\frac 1{n^{2-x}}$.