Determine con-/divergence of $\sum\limits_{n=1}^{\infty}\frac{1+2+\cdots+n}{2^n}$ Problem:
Determine if $\sum\limits_{n=1}^{\infty}\frac{1+2+\cdots+n}{2^n}$ converges or diverges.
My attempt:
I'm having a hard time with this one.


*

*Trying the ratio test, I'm unable to simplify the expression

*Trying the root test, I get the nth root of a sum, which I don't know how to simplify

*Trying the integral test, I'm having a hard time performing the actual integration

*Trying the (limit) comparison test, I can't find any reasonable expression to compare it with, but that might be due to the fact that I've been working on sequences like this for less than a day


Any help appreciated!
 A: Notice that $1+\dots+n\leqslant n^2$, then you can conclude convergence with the ratio test.
A: We can actually evaluate the sum using the binomial theorem (and negative binomial coefficients):
$$
\begin{align}
\sum_{n=1}^\infty\frac{1+2+3+\cdots+n}{2^n}
&=\sum_{n=1}^\infty\frac{\binom{n+1}{2}}{2^n}\\
&=\sum_{n=1}^\infty\frac{\binom{n+1}{n-1}}{2^n}\\
&=\sum_{n=1}^\infty(-1)^{n-1}\frac{\binom{-3}{n-1}}{2^n}\\
&=\frac12\sum_{n=0}^\infty(-1)^n\frac{\binom{-3}{n}}{2^n}\\
&=\frac12\left(1-\frac12\right)^{-3}\\[9pt]
&=4
\end{align}
$$
A: define $f(x)=\frac{4x}{(2-x)^3}$ write a Taylor's series to obtain$$f(x)=\sum_{n=1}^{\infty}\frac{n(n+1)}{2^{n+1}}x^n$$which converges on $|x|<2$, therefore your sum is $f(1)=4$.
A: Hint:You can write the numerator $1+2+\dots+n$ as $\frac{n(n+1)}{2}$
A: One more way is by using infinite GP, denote the required sum as 
$S(x) =\sum_{k=0}^{\infty} \frac{k(k+1) x^k}{2}$, we want $S(1/2)$
$\sum_{k=0}^{\infty} x^k =\frac{1}{1-x}, |x|<1$;
d.w.r.t x on both sides to get
$\sum_{k=0}^{\infty} k x^k =\frac{x}{(1-x)^2}=f(x)$
Again d.w.r.x on both sides to get
$\sum_{k=0}^{\infty} k^2 x^k = x f'(x).$
$S(x)=[f(x)+x f'(x)]/2=\frac{x}{(1-x)^3}.$
Finally we get $S(1/2)=4.$
