Telescoping sums
- Thread starter GetReal
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Okay, having proved the given identity, then we know:
[MATH]\frac{k}{7^k}=\frac{7}{36}\left(\frac{6k+1}{7^{k}}-\frac{6(k+1)+1}{7^{k+1}}\right)[/MATH]
And so, the sum \(S\) we are asked to simplify becomes:
[MATH]S=\frac{7}{36}\left(\sum_{k=0}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=0}^{n-1}\left(\frac{6(k+1)+1}{7^{k+1}}\right)\right)[/MATH]
Let's re-index the second sum as follows:
[MATH]S=\frac{7}{36}\left(\sum_{k=0}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=1}^{n}\left(\frac{6k+1}{7^{k}}\right)\right)[/MATH]
Now, strip off the first term from the first sum and the last term from the second sum:
[MATH]S=\frac{7}{36}\left(1+\sum_{k=1}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=1}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\frac{6n+1}{7^{n}}\right)[/MATH]
Can you proceed?
[MATH]\frac{k}{7^k}=\frac{7}{36}\left(\frac{6k+1}{7^{k}}-\frac{6(k+1)+1}{7^{k+1}}\right)[/MATH]
And so, the sum \(S\) we are asked to simplify becomes:
[MATH]S=\frac{7}{36}\left(\sum_{k=0}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=0}^{n-1}\left(\frac{6(k+1)+1}{7^{k+1}}\right)\right)[/MATH]
Let's re-index the second sum as follows:
[MATH]S=\frac{7}{36}\left(\sum_{k=0}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=1}^{n}\left(\frac{6k+1}{7^{k}}\right)\right)[/MATH]
Now, strip off the first term from the first sum and the last term from the second sum:
[MATH]S=\frac{7}{36}\left(1+\sum_{k=1}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\sum_{k=1}^{n-1}\left(\frac{6k+1}{7^{k}}\right)-\frac{6n+1}{7^{n}}\right)[/MATH]
Can you proceed?