Evaluate the sum of series

Find the sum of following series: $\frac<1>+\frac<1>+\frac<1>+. $ After some arrangement, I got below step: $\frac<1>+\frac<1>+\frac<1>+. =\sum\limits_^<\infty>\frac<1>- \sum\limits_^<\infty>\frac<1> $ Now, I have no idea how to find this difference of two series. Please help me. Thanks in advance.

Siddhant Trivedi asked Dec 29, 2015 at 15:40 Siddhant Trivedi Siddhant Trivedi 2,318 5 5 gold badges 29 29 silver badges 42 42 bronze badges $\begingroup$ Expand $$\ln(1+x)$$ $\endgroup$ Commented Dec 29, 2015 at 15:45

$\begingroup$ The series on the left converges, and the two series on the right do not, so that approach may not be useful. Your series is $\frac<1>-\frac<1>+\frac<1>-\frac<1>+\cdots$. Now have you seen something like this before? $\endgroup$

Commented Dec 29, 2015 at 15:53 $\begingroup$ @lab battacharjee $ 1-\log 2$ $\endgroup$ Commented Dec 29, 2015 at 15:55 $\begingroup$ how after some (re-)arrangement you reached the summation difference? $\endgroup$ Commented May 8, 2021 at 15:14

2 Answers 2

$\begingroup$ answered Dec 29, 2015 at 15:53 Start wearing purple Start wearing purple 53.4k 13 13 gold badges 165 165 silver badges 225 225 bronze badges $\begingroup$ I got the answer before your post. Thanks $\endgroup$ Commented Dec 29, 2015 at 16:03 $\begingroup$ How did you move from the telescoping expression to the alternating one? $\endgroup$ Commented Dec 29, 2015 at 16:15

$\begingroup$ @zz20s The $k$th term in the first expression is the sum of $2k$th and $(2k+1)$th terms in the second (e.g. $\frac12-\frac13=\frac2 +\frac3$). $\endgroup$

Commented Dec 29, 2015 at 16:39 $\begingroup$ That makes sense, thank you. $\endgroup$ Commented Dec 29, 2015 at 16:40 $\begingroup$ @zz20s I converted each terms appear in sum to factorial form. $\endgroup$ Commented Dec 30, 2015 at 6:15 $\begingroup$

The general term for the above series is 1/(2r)(2r+1). This can be broken into partial fractions. Write the numerator as 2r+1-2r and thus this general term can be written as 1/(2r+1) - 1/(2r). sum the terms from r=1 to n, and then take the answer in a limit n approaching infinity.

answered Dec 29, 2015 at 15:50 $\begingroup$ Please use MathJax in your posts. $\endgroup$ Commented Dec 29, 2015 at 15:57

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