Series equation

**Lisa91** · January 15th 2013, 11:00 AM

so I get $\frac{4}{3} S$ , which is great but how the S looks? How to make all idices the same?

**TheSaviour** · January 15th 2013, 11:02 AM

Originally Posted by Lisa91

so I get $\frac{4}{3} S$ , which is great but how the S looks? How to make all idices the same?

Indexes have no impact on the outcome of the sum so you call all of them k or r or m or f or g or whatever you want!

**Lisa91** · January 15th 2013, 11:15 AM

Well... We have $3 \sum_{m=1}^{\infty} \frac{1}{m3^{m}} - 1-\frac{1}{6} +\frac{1}{3} \sum_{k=1}^{\infty} \frac{1}{k3^{k}}-2\sum_{n=1}^{\infty} \frac{2}{n3^{n}}$ . Then we get $\frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n3^{n}} -\frac{7}{6}$ which is wrong. We should get $-\frac{1}{2}$ .

**TheSaviour** · January 15th 2013, 11:45 AM

Originally Posted by Lisa91

So, I got $\sum_{m=1}^{\infty} \frac{3}{m3^{m}} - \frac{7}{6} + \frac{1}{3}\sum_{k=1}^{\infty} \frac{1}{k3^{k}} - \sum_{n=1}^{\infty} \frac{2}{n3^{n}}$ How to combine them all? I don't know how to cope with those $2,3, \frac{1}{3}$ . For instance, if I go back to n, I get n=0 in the first term, which is not what I want...

Here's the mistake. The last one should have began from n = 2 (since we didn't do anything to it). So write it from n = 1 by subtracting the value at n = 1.

**Lisa91** · January 15th 2013, 12:06 PM

Thank you all so much! Now it's ok

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