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how to show that $$\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$$? Can I use the Alternating Series test and how?

Darío A. Gutiérrez
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Andrej
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3 Answers3

5

$$\dfrac{3n-1}{n(n+1)}=\dfrac{3(n+1)-4}{n(n+1)}=\dfrac3n-4\left(\dfrac1n-\dfrac1{n+1}\right)=\dfrac4{n+1}-\dfrac1n$$

This can also be achieved by Partial Fraction Decomposition $$\dfrac{3n-1}{n(n+1)}=\dfrac An+\dfrac B{n+1}$$

Now $\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$ for $-1\le x<1$ by

What is the correct radius of convergence for $\ln(1+x)$?

lab bhattacharjee
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HINT

We have that

$$(-1)^{n}\dfrac{3n-1}{n^2 + n} =(-1)^{n}\frac1n\dfrac{3n+3-4}{n + 1} =3\frac{(-1)^{n}}n-4(-1)^{n}\dfrac{1}{n(n + 1)}=\ldots$$

and since by telescoping

$$\dfrac{1}{n(n + 1)}=\frac1{n}-\frac1{n+1}$$

we obtain

$$-\frac{(-1)^{n}}n+4\dfrac{(-1)^{-1}}{n + 1}$$

then recall that by alternating harmonic series

  • $\sum_{n=1}^\infty \frac{(-1)^{n}}n=\ln 2$
user
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    What is the next step? – lab bhattacharjee Oct 22 '18 at 13:27
  • @labbhattacharjee I was thinking to use the alternating harmonic series for the first term and telescoping for the second one. – user Oct 22 '18 at 13:29
  • @labbhattacharjee I've added some detail more, it should be sufficient now as a hint I think. – user Oct 22 '18 at 13:35
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    -1 for posting a poor answer, and only in response to a comment, improving it after posting, rather than before submitting it in the first place (not to mention you supplement your post with work from another answer). – amWhy Oct 22 '18 at 13:36
  • @amWhy I agree that maybe my forst hint wasn't to much cryptically formulated at first but my aim was to let some work to the asker. Of course I would at lest give a reference to the alternating harmonic series which is necessary to conclude the derivation according to my suggestion. – user Oct 22 '18 at 13:41
  • @amWhy Often we say: "do not give full solution" but I see that some times we say: "not enought hint was given for the solution". Belive me it is not a simple task to satisfy all the requests! – user Oct 22 '18 at 13:42
  • @amWhy In particular when the question are not properly posted and the asker has not shown completely his effort a small hint is tha best answer we can give. Duplicates, complete answers and comments with complete answers are the worst thing to do for the benefit of the community since that stimulate the proliferation of such kind of questions. – user Oct 22 '18 at 13:45
  • @amWhy Errata corrige for my first comment to you: "I agree that maybe my first hint was to much cryptically formulated..." – user Oct 22 '18 at 13:47
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$$S_1= \sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n}$$ $$\dfrac{3n-1}{n^2 + n} = \dfrac{3-\frac{1}{n}}{n + 1} = \left(\frac{4}{n+1} - \frac{1}{n}\right)$$ So \begin{align} S_1 &= \sum_{n=1}^{\infty}(-1)^{n}\left(\frac{4}{n+1} - \frac{1}{n}\right)\\ &= 4\underbrace{\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n+1}}_{s_2} - \underbrace{\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}}_{-\ln(2)} \end{align} We know, then $$\ln(1 + x) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k} = x - \frac{x^2}{2} + \frac{x^2}{3} - \frac{x^2}{4} + \cdots$$ and for $x = 1$ we have $$\ln(1 + 1) = \ln(2) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{1^k}{k} = 1 \underbrace{- \frac{1^2}{2} + \frac{1^2}{3} - \frac{1^2}{4} + \cdots}_{s_2} = 1- s_2$$

Which implies $$s_2 = \ln(2) -1$$ Therefore \begin{align} S_1 &= 4(s_2) - (-\ln(2))\\ &= 4(\ln(2) -1) - (-\ln(2))\\ &= 4\ln(2) -4 + \ln(2)\\ &= 5\ln(2) -4\\ &= \ln(32) -4\\ \end{align}
See:

the sum: $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$ using Riemann Integral and other methods

Darío A. Gutiérrez
  • 4,007