# Sum of conditionally convergent series

The series $\sum_{n=1}^{\infty}2^{-n}$ is absolutely convergent. It converges to $\frac{1}{1-\frac{1}{2}}$.

Series $\sum_{n=1}^{\infty}(-1)^{n}(2^{-n}+n^{-1})$ is convergent, but only conditional convergent. Does this mean that it does not converge to any specific value like the geometric series? I am not quite sure that I am fully getting what conditional convergence means.

• A consequence of conditional convergence is that you can rearrange the terms of the series to make it not converge, and in fact to make it converge to whatever value you please. An absolutely convergent series converges to the same value no matter how it is rearranged. Jun 16, 2018 at 18:43

Of course it converges to a specific value; otherwise, it wouldn't be convergent. A simpler example is $\sum_{n=1}^\infty\frac{(-1)^{n-1}}n$; it converges conditionally to $\log2$.
This is because if $a_n=(-1)^{n}(2^{-n}+n^{-1})$$\sum_{n=1}^{\infty}a_n=-\dfrac{1}{3}-\ln 2$$but$$\sum_{n=1}^{\infty}|a_n|>\sum_{n}\dfrac{1}{n}=\infty$$Also Bernhard Riemann proved that a conditionally convergent series of real numbers may be rearranged to converge to any value at all, including ∞ or −∞ [Wikipedia] $$\sum_{n=0}^{\infty}{f(n)}$$ is described as absolutely convergent IF $$\sum_{n=0}^{\infty}{|f(n)|}$$ converges. Think:$|x|$means the absolute value of$x$. Conditional convergence is simply when this criteria is not met.$\sum_{n=1}^{\infty}(-1)^{n}(2^{-n}+n^{-1})$converges to$-\frac13-\log_e(2)\approx -1.02648$when read as the limit of the partial sums of$-\left(\frac1{2^1}+\frac11\right)+\left(\frac1{2^2}+\frac12\right)-\left(\frac1{2^3}+\frac13\right)+\left(\frac1{2^4}+\frac14\right)-\left(\frac1{2^5}+\frac15\right)+\left(\frac1{2^6}+\frac16\right)- \cdots $but if you reorder, for example to two negative terms followed by one positive so still retaining all the terms in the infinite series, as in$-\left(\frac1{2^1}+\frac11\right)-\left(\frac1{2^3}+\frac13\right)+\left(\frac1{2^2}+\frac12\right)-\left(\frac1{2^5}+\frac15\right)-\left(\frac1{2^7}+\frac17\right) +\left(\frac1{2^4}+\frac14\right)-\cdots $then the limit of the partial sums is closer to$-1.373$Compare this with$\sum_{n=1}^{\infty}(-1)^{n}(2^{-n})$which is absolutely convergent as$\sum_{n=1}^{\infty}|(-1)^{n}(2^{-n})|=1$is finite: you then know that the limits of the partial sums of$-\frac1{2^1}+\frac1{2^2}-\frac1{2^3}+\frac1{2^4}-\frac1{2^5}+\frac1{2^6}-\cdots$and of$-\frac1{2^1}-\frac1{2^3}+\frac1{2^2}-\frac1{2^5}-\frac1{2^7}+\frac1{2^4}-\cdots$will converge to the same value, in fact$-\frac13\$