(Hi)Hi.

The problem is giving me a hard time basically because i just can't seem to figure out how to solve for it.

The problem is $\displaystyle \sum(-1)^n\int1/2^x$

the limits on sigma are n=1 to infinity and for the integral its n and n+1

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- Apr 26th 2009, 09:49 AM[s]arahconfussing infinite series problem
(Hi)Hi.

The problem is giving me a hard time basically because i just can't seem to figure out how to solve for it.

The problem is $\displaystyle \sum(-1)^n\int1/2^x$

the limits on sigma are n=1 to infinity and for the integral its n and n+1 - Apr 26th 2009, 10:09 AMJhevon
As in $\displaystyle \sum_{n = 0}^\infty (-1)^n \int_n^{n + 1}\frac 1{2^x}$??

Where are you getting stuck? just do the integral, you get

$\displaystyle \sum_{n = 0}^\infty (-1)^{n+1} \frac 1{2^x \ln 2} \bigg|_n^{n + 1}$

now, plug in the limits of integration, factor out the ln(2), and you can split the sum into two geometric series