# Checking convergence of series

My task is to check whether the series $$\sum_{n=1}^\infty {\frac{(-2)^n+n^2}{n2^n}}$$ converges or not. So I tried expanding the fraction like that(should equal to original form): $$\sum_{n=1}^\infty {(-1)^n\frac{2^n}{n2^n}+\frac{n^2}{n2^n}}=\sum_{n=1}^\infty {(-1)^n\frac{1}{n}+\frac{n}{2^n}}$$

Then I thought about that the first addend does converge because of Leibniz's rule and the second addend converges because $2^n$ grows much faster than $n$.

But I'm not sure(and rather thinking this form of argumentation is wrong) if I can take addends of a series and prove convergence for each addend instead of proving convergence for the whole series.

I would appreciate if someone can point me to the right direction.

• You are using as an upper limit the "dummy index" $n$. Dec 1, 2015 at 4:50
• It is a theorem that if $\sum_0^\infty a_n$ and $\sum_0^\infty b_n$ converge, then $\sum_0^\infty (a_n+b_n)$ converges. Dec 1, 2015 at 4:50
• @Dr.MV thank you, corrected it, it should be infinty Dec 1, 2015 at 4:55
• You're welcome. My pleasure. And @AndréNicolas has answered the question. Are you good to go now? Dec 1, 2015 at 4:57
• Yep, good to go now :) Dec 1, 2015 at 4:59

The 1st one is $-\ln 2$ (Taylor series).
The 2nd one is $$f(x)=\sum_{n=1}^{\infty}n\cdot x^n$$ $$\int \frac{f(x)}{x}dx=\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$$ $$\therefore f(x)=x\cdot\left(\frac{x}{1-x}\right)'=\frac{x}{(1-x)^2}$$ $$\sum_{n=1}^{\infty}n\cdot\left(\frac12\right)^n=\frac{\frac12}{(1-\frac12)^2}=2$$ So the total sum is $2-\ln2$.
To control if the series: $$\sum_{n=1}^\infty {(-1)^n\frac{1}{n}+\frac{n}{2^n}}$$ you can use the Leibniz's rule. First you have to demonstrate that $$-\frac {1}{n}+\frac {n}{2^n}<0$$ for every $n$. Indeed you can note that rewriting the expression $$-\frac {2^n+n^2}{n\cdot 2^n}<0$$. Therefore now you know that for $n\to \infty$ $$\frac {1}{n}+\frac {n}{2^n}\to 0$$ therefore you have to demonstrate $$\frac {1}{n}+\frac {n}{2^n}$$ is decreasing proving that the derivate of $f(x)=\frac {1}{x}+\frac {x}{2^x}<0$ after a value of $x$.