Preferred way in calculating sum (calculus) What is the preferred way of calculating the value of a summation?
Is it Riemann's Sum?
For example:
$$\sum_{n=1}^{\infty} { (-1)^n \frac{3^n}{n4^n} }$$
 A: Recall that for $\vert x \vert < 1$, we have
$$\sum_{n=1}^{\infty} \dfrac{x^n}n = - \log(1-x)$$
A quick way to see this is to note that for $\vert t \vert < 1$, we have
$$\sum_{n=1}^{\infty} t^{n-1} = \dfrac1{1-t} \tag{$\star$}$$
Integrating $(\star)$ from $0$ to $x$ (where $\vert x \vert < 1$), we get that
$$-\log(1-x) = \int_0^x\dfrac{dt}{1-t} = \int_0^x \sum_{n=1}^{\infty} t^{n-1} dt= \sum_{n=1}^{\infty}  \int_0^x t^{n-1} dt = \sum_{n=1}^{\infty} \dfrac{x^n}n\tag{$\perp$}$$
which gives us what we want.
A: Most preferred ways of computing sums like those boil down to finding a way (vis differentiation, integration etc.) to reduce them to a geometric series, and then working from there.
Consider
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n} x^n$$
where $|x| < 1$.  This is actually a well-known sum, but let's go from here as I described.  If we differentiate with respect to $x$, we get
$$\frac{d}{dx} \sum_{n=1}^{\infty} \frac{(-1)^n}{n} x^n = \sum_{n=1}^{\infty} (-1)^{n} x^{n-1} = -\frac{1}{1+x}$$
We may then integrate to get
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n} x^n = -\int \frac{dx}{1+x} = \log{\left( \frac{1}{1+x}\right)}$$
where the integration constant is zero as the sum evaluated at zero is zero.
Your sum has $x=3/4$.
A: There are many ways.  In this case you might recognize the Taylor series for $\log (1+x)=\sum_{i=1}^\infty \frac {(-1)^{i+1}x^i}i$, substitute in $x=\frac 34$ and multiply by $-1$ to get $-\log (\frac 74)$
A: Hint (without exaplanation/justification, to fill in):
$$\forall\,x\in\Bbb C\;,\;|x|<1\;:\;\;\frac{1}{1+x}=\sum_{k=0}^\infty(-1)^nx^k\implies$$
$$\log(1+x)=\int\left(\sum_{k=0}^\infty(-1)^kx^k\right)dx=\sum_{k=0}^\infty(-1)^n\int x^k\,dx=\sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k+1}\ldots$$
A: To answer the question behind this example, there is little in the way of "general results". Just like there are different ways of reducing given integrals to a handful of typical examples that can be solved, there are some techniques (eerily reminiscent to the integral case, both are sums) to reduce sums to known forms. The answers given hint at one of the more important of them: Try to represent your sum as some mangling of a known (here geometric) series. There are specific techiques for some types of sums (see for example Petkovsek, Wilf, Zeilberger's "A = B", and also Wilf's "generatingfuctionology") Then there are all sort of neat (and also some hair-raising) tricks to tame specific other cases, and a vast collection of sums that just can't be reduced to anything neat. The solution then is to resort to approximate methods of one sort or the other, and on those there is a vast amount of work too.
