# sum $1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + ...$ [duplicate]

find the sum if the following series converges:

$$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + ...$$

my attempt:

The series can be grouped into difference of a series of odd terms and a series of even terms.. but how to find sum?

• Can you elaborate on the pattern of the denominators? It's not obvious to me. Aug 14, 2015 at 9:50
• @Hirshy No, not really a duplicate. As I see it, this question specifically asks for the limit $$\lim_{n \to \infty}\sum_{i = 1}^n \frac{1}{4i-3} + \frac{1}{4i-1} - \frac{1}{2i}$$ Aug 14, 2015 at 10:00
• WolframAlpha says it sums to $\frac{\ln 8}{2} = \frac{3\ln 2}{2}$ Aug 14, 2015 at 10:05
• A typical group of three terms is $\frac 1{4n-1}-\frac 1{2n}+\frac 1{4n+1}=\frac 1{4n-1}-\frac 1{4n}-\frac 1{4n}+\frac 1{4n+1}=\frac 1{4n(4n-1)}-\frac 1{4n(4n+1)}=\frac 2{(4n-1)4n(4n+1)}$ Aug 14, 2015 at 10:21
• Can you express it as difference of 2 Riemann sums? Aug 14, 2015 at 10:22

Let $L=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\dots \tag{1}$

$(1)$ converges by Alternating series test, so $L$ is finite

Go through this link to convince yourself that $L=\ln(2).$

Now multiply $L$ by $\frac{1}{2}$, we get $$\frac L2=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}\dots \tag 2$$

Now adding $(1)$ and $(2)$, we get

$$\frac{3}2L=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}\dots \tag 3$$

Your question ends here as you have found the required sum as $\frac{3}2(\ln(2)) \approx 1.0397$

Now notice that $(3)$ is just a rearrangment of $(1)$, in a way, we list the first two +ve terms followed by first -ve term from $(1)$ to make up first three terms of $(3)$ , and then next two +ve terms followed by second -ve term, and so on.

Thus we see that rearrangments of a series are not necessarily to converge on the same limit. And thus, an interesting question pops up- "when does rearrangements converge to same limit?"