0
$\begingroup$

I want to calculate the series $$\sum_{n=1}^m\frac{1}{2n+1}$$ I have found out how to do this with $$\sum_{n=1}^m\frac{1}{2n-1}$$ but I cannot find this. Please help!

  • 0
    the first one is just the second one, when $n$ is substituted by $n+1$, plus an extra term2017-01-03
  • 2
    How do you do the second sum? Any way other than just adding them up?2017-01-03

2 Answers 2

2

If you know how to deal with the second, just note that the two are very stricly related, as: $$\sum_{n=1}^m\frac{1}{2n-1}=1+\sum_{n=1}^m\frac{1}{2n+1}-\frac{1}{2m+1}$$

0

Let $S_1$ be the first sum and $S_2$ be the second sum. Then $$ S_1 = S_2 - 1+\frac{1}{2m+1} $$