Does $\sum_{n=1}^{\infty} (-1)^n \cdot \frac{n}{2^n} $ diverge or converge? I want to prove that $\sum_{n=1}^{\infty} (-1)^n \cdot  \frac{n}{2^n}$ converges.
I was trying the ratio test:
$\sum_{n=1}^{\infty} a_n $ converges if $ q:=\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1$ and diverges if $q >1$.
So we get: $\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} = \lim_{n\to\infty} \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \lim_{n\to\infty} \frac{n2^n+2^n}{n2^{n+1}}$.
How do you calculate the limit of this function? I'm currently practicing for an exam and the solution just used this step:
$\lim_{n\to\infty} \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} = \frac{1}{2}(1+\frac{1}{n}) = \frac{1}{2} < 1$ therefore convergent. However I really don't see how you transform $\frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}}$ to $\frac{1}{2}(1+\frac{1}{n})$.
 A: When you use the ratio test, reorganize the fraction into two fractions made of similar pieces. So when you have $\frac{n+1}{2^{n+1}}\cdot\frac{2^n}{n}$, move things around to get $\frac{n+1}{n}\cdot\frac{2^n}{2^{n+1}}$. Since similar expressions are now together, it makes simplifying easier. In this case you can rewrite it as $(1+\frac{1}{n})\cdot\frac12$. (If the first factor had more in the denominator than just $n$, so you could split it into 2 fractions, you could use l'Hospital's rule, or there are general rules for finding limits of rational functions.)
This kind of regrouping is also crucial when your series has factorials in it.
A: When $|x|<1$ the geometric series $$(1+x)^{-1}=\sum_{k=0}^{\infty} (-x)^k$$ converges
. Differentiating it w.r.t $x$, we get
$$-(1+x)^{-2}=-(-x)^{-1}\sum_{k=0} k(-x)^k \implies S=\sum_{k=1}^{\infty} (-1)^k x^k=-x(1+x)^{-2}$$
Let $x=1/2$, then $$S=-2/9<\infty.$$
So the series converges.
A: This is an alternating series, i.e., the terms alternate between positive and negative.
Also, the general term abslute value $|a_n| = \frac{n}{2^n} \to 0$ monotonically, as $n\to \infty$.
Then, by the Alternating series test, we have that the series converge.
A: Take $a_n=\frac n{2^n}$, then we have that $a_n\geq 0$ and
$$a_n-a_{n+1}=\frac{n-1}{2^{n+1}}\text{ for all }n\in\mathbb{N}$$
$$\implies a_n\geq a_{n+1}\text{ for all }n\in\mathbb{N}$$
also $\lim_{n\to\infty}a_n=0$. Hence by Alternating series test
,
$$\sum_{n=1}^\infty(-1)^na_n\text{ connverges}$$
