I need to evaluate the sum: can someone help?
The series is as follows: $$ \frac 14 - \frac {1}{2(4)^2} + \frac{1}{3(4)^3} - \cdots + \frac{(-1)^{(x+1)}}{x(4)^x} $$
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This is $log(1+1/4)$. You can use the power series of logarithms to obtain it.
Amr
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I don't think it is ... becuase log(1+1/4) is not ((-1)^(n+1)*x^n)/n – mc1 Nov 22 '12 at 19:16
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@mc1 Check out http://en.wikipedia.org/wiki/Taylor_series and read the section on the Taylor series for log – Adam Cross Nov 22 '12 at 19:39
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We know, $$\sum_{0\le s\le n-1}y^s=\frac{y^n-1}{y-1}$$
Integrating either sides wrt $y$, we get $$\sum_{1\le r\le n}\frac{y^r}r=\int \left(\frac{y^n-1}{y-1}\right) dy$$
If $\sum_{1\le r\le n}\frac{y^r}r=S(y,n),$
$\frac 14 - \frac {1}{2(4)^2} + \frac{1}{3(4)^3} - \cdots + \frac{(-1)^{(x+1)}}{x(4)^x}=-S(-\frac14, x+1)$
The formula for the general summation is explained here.
lab bhattacharjee
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