GRE Problems - Variants

Solution Whenever an infinite series involving a power of \(x\) is involved, you should think of the geometric series formula, that we review here below: \[\sum_{n=0}^\infty x^n =\frac{1}{1-x} \mbox{ for } |x|\lt 1.\] A vast number of problems involving infinite series are done by using the geometric series. We will also need a standard manipulation technique for sums, the shift of the index of summation. The idea is simple: we can describe a sum such as \[a_1+a_2+a_3+\cdots\] either as the sum of \(a_n\) as \(n\) ranges from \(1\) to \(\infty\), or the sum of \(a_{n+1}\) as \(n\) ranges from \(0\) to \(\infty\), or the sum of \(a_{n-1}\) as \(n\) ranges from \(2\) to \(\infty\), and so on. Translating this to the summation notation, it means that \[\sum_{n=1}^\infty a_n=\sum_{n=0}^\infty a_{n+1} =\sum_{n=2}^\infty a_{n-1}=\cdots = \sum_{n=k}^\infty a_{n+1-k}.\] So we see that we can change the starting value of the index from \(1\) to \(k\), as long as we replace the terms \(a_n\) we are summing with \(a_{n+1-k}\). If the starting value is \(0\), we can increase it to \(k\) if we replace \(a_n\) with \(a_{n-k}\). So for example for the geometric series we have \[\sum_{n=0}^\infty x^n =\sum_{n=1}^\infty x^{n-1}.\] In this problem, the sum doesn't quite look like a geometric series. But it gets closer after we take the derivative: \begin{eqnarray*} f'(x)&=&\sum_{n=1}^\infty \frac{(-1)^n}{n} \frac{d}{dx}x^{2n}\\ &=& \sum_{n=1}^\infty \frac{(-1)^n}{n} 2nx^{2n-1}\\ &=& 2\sum_{n=1}^\infty (-1)^n x^{2n-1} \end{eqnarray*} The next thing to do is try to isolate an \(n\)-th power, so that we can do some sum manipulations. So we re-write \[x^{2n-1}=\frac{\left(x^2\right)^n}{x}.\] We can also combine the \(n\)-the powers \((-1)^n\) and \(\left(x^2\right)^n\) into a single one: \((-1)^n \left(x^2\right)^n=\left(-x^2\right)^n\) and we find \begin{eqnarray*} f'(x)&=& \frac{2}{x}\sum_{n=1}^\infty \left(-x^2\right)^n \end{eqnarray*} All we need to do now is shifting the starting value by \(1\) to get a geometric series for the variable \(\left(-x^2\right)\): \begin{eqnarray*} f'(x)&=& \frac{2}{x}\sum_{n=0}^\infty \left(-x^2\right)^{n+1}\\ &=&\frac{2}{x}(-x^2)\sum_{n=0}^\infty \left(-x^2\right)^n\\ &=&-2x\frac{1}{1-(-x^2)}=-\frac{2x}{1+x^2}. \end{eqnarray*} Note that it would be easy to do the integral of this answer, using the substitution \(u=1+x^2\). So this method allows us to find a simple formula not only for \(f'(x)\) but also for the original \(f(x)\).