Question 1018961
Prove that:
<pre>
{{{sum((2k-1^"")^""^""^"",k=1,n)}}}{{{""=""}}}{{{sum( ((n^2-1)/n^2)^(k-1),k=1,infinity)}}}

The left side is the sum of the arithmetic series:

1 + 3 + 5 + ... + 2n-1

The formula for this sum is

{{{S[n]}}}{{{""=""}}}{{{expr(n/2)(a[1]+a[n])}}}

where a<sub>1</sub> = 1 and a<sub>n</sub> = 2n-1 

{{{S[n]}}}{{{""=""}}}{{{expr(n/2)(1+(2n-1)^"")}}}{{{""=""}}}{{{ expr(n/2)(1+2n-1)}}}{{{""=""}}}{{{expr(n/2)(2n)}}}{{{""=""}}}{{{ n^2}}}

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Now we see if we can show that the right side is also 
equal to n<sup>2</sup>.


The right side is the sum of an infinite geometric series

The formula for this sum is 

{{{S[infinity]}}}{{{""=""}}}{{{a[1]/(1-r)}}}

where a<sub>1</sub> = 1 and {{{r }}}{{{""=""}}}{{{ (n^2-1)/n^2}}}{{{""=""}}}{{{n^2/n^2-1^""/n^2}}}{{{""=""}}}{{{1-1^""/n^2}}} 

{{{S[infinity]}}}{{{""=""}}}{{{1^""/(1-(1-1/n^2))}}}{{{""=""}}}{{{1^""/(1-1+1/n^2)}}}{{{""=""}}}{{{1^""/(1/n^2)}}}{{{""=""}}}{{{n^2}}}

So the right side also equals n<sup>2</sup>. 
That proves they are equal.

Edwin</pre>