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If one lets $G$ denote the series

$\hspace{2in}$$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdot\cdot\cdot$,

then it is easy to see that because $2G$ produces

$\hspace{2in}$$2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{80} + \cdot\cdot\cdot$,

which is exactly $2 + G$, then $G = 2$ because $2G = 2 + G$; however, if one then applies this to the series $H$

$\hspace{2in}$$1 + 2 + 4 + 8 + 16 + \cdot\cdot\cdot$

$2H$ gives

$\hspace{2in}$$2 + 4 + 8 + 16 + 32$,

which is exactly $H - 1$, but this then means $H = -1$, which is clearly not the case. Is it true that this trick only works with convergent series? Is there a case where this trick does not work for a convergent series or where it does work for a divergent series? Any insight is much appreciated. Thank you.

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The trick works for convergent geometric series. There is no obvious way to modify the trick to evaluate the sum of the convergent series $\sum_{n=1}^\infty \frac{1}{n^2}$. But once we know $\sum_1^\infty \frac{1}{n^2}$, the trick can be adapted to find, for example, $\sum_1^\infty \frac{1}{(2n+1)^2}$. –  André Nicolas Mar 2 '13 at 7:00
    
I'll tell you more $$ H = 1-1+1-1+1-1 + \ldots \\ -H = -1+1-1+1-1+1 -\ldots = H-1 \\ 2H=1 \\ H=\frac 12 $$ –  Kaster Mar 2 '13 at 7:06
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you need to be careful with these kind of questions where you do a substitution - the very first thing you need to do is to show that a limit in fact exists, and only then can you do the trick of substitution. –  Vincent Tjeng Mar 2 '13 at 7:25
    
Also, en.wikipedia.org/wiki/… –  lab bhattacharjee Mar 2 '13 at 7:30
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1 Answer 1

The trick works also for the divergent series: $$ H - 1 = 2H $$ has also $+\infty$ as a solution. However you cannot subtract $H$ to each side if you don't know that $H$ is finite.

It is true that when a series has a limit (finite or not) then you can multiply your series by a constant, and the limit of the new series gets multiplied by the constant. You can also sum two series term by term, and if you don't have an indeterminate operation (like $\infty - \infty$), then the result is correct (you also need to know that the series do converge, of course). Also you can shift a series like you did, and the result gets shifted by a constant.

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