# Finding the sum of a series $\frac{1}{1 \cdot 2}-\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}-\frac{1}{4 \cdot 5}+…$

The value of:

$$\frac{1}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \cdots$$

is

(A) between 0 and 1/4.

(B) between 1/4 and 1/3.

(C) between 1/3 and 1/2.

(D) between 1/2 and 1.


I was looking for a convenient way to tackle it. Can someone point me in the right direction? Thanks in advance for your time.

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Which problem would you like someone to solve? –  Ron Gordon Jan 21 '13 at 14:06
sorry for the wrong post.I have edited it. –  learner Jan 21 '13 at 14:09
Isn't it time for you to have learned TeX, rather than posting images? –  Thomas Andrews Jan 21 '13 at 14:09
Does the above really mean $1.2=\frac{12}{10}$ or does it mean $1\times 2$? –  Thomas Andrews Jan 21 '13 at 14:13
It means $1\times 2$ –  learner Jan 21 '13 at 14:19

This is an alternating series with monotonely (ignoring the signs) decreasing terms. The proof of Leibniz' test shows that the value of the series $s$ lies between two consecutive partial sums.

Hence $$\frac{1}{2} > s > \frac{1}{2}-\frac{1}{6} = \frac{1}{3}.$$

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Consider

$$\log{(1+x)} = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} x^k$$

$$\int_0^1 dx \: \log{(1+x)} = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \left [ \frac{x^{k+1}}{k+1} \right ]_0^1 = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k (k+1)}$$

But

$$\int_0^1 dx \: \log{(1+x)} = \left [ (1 + x) (\log{(1+x)}-1) \right ]_0^1 = 2 \log{2} - 1$$

This is the result you seek.

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Sorry but I do not want to recall the value of log(2) to choose between (A), (B), etc. –  Did Jan 21 '13 at 14:36
Could you explain why this integral is the sum of the OP's series. –  Peter Phipps Jan 21 '13 at 14:37
@did: fixed the error before you posted. –  Ron Gordon Jan 21 '13 at 14:41
It seems unlikely that the OP is that familiar with the Taylor series for $\log(1+x)$ and its convergence properties (you need Abel's theorem to guarantee that the series on line 1 actually converges to $\log 2$ when $x = 1$ and something similar for the integrated series). –  mrf Jan 21 '13 at 14:50
No problem. Let us move on, there is plenty of mathematics out there, just waiting to give us joy and pleasure... –  Did Jan 21 '13 at 15:17
It's same as $(1-\frac{1}{2})-(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})-(\frac{1}{4}-\frac{1}{5})+\cdots=1+2(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots)$
$=1+2(\ln2-1)=2\ln2-1\approx 0.3862$