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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.

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  • $\begingroup$ 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. $\endgroup$ Jun 16, 2018 at 18:43

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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 $\log2$.

Being conditionally convergent simply means that the series of the absolute values diverges.

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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]

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$$\sum_{n=0}^{\infty}{f(n)}$$ 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.

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$\sum_{n=1}^{\infty}(-1)^{n}(2^{-n}+n^{-1})$ 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 $-\frac13$

This is the key point: if a series is absolutely convergent (i.e. the sum of the absolute values converges to a finite value) then the sum of the actual values will also be finite and will not depend on the order of the terms; if a series is conditionally convergent but not absolutely convergent then changing the order of the actual values can change the result and even make the magnitude of the limit of the partial sums arbitrarily large or infinite

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