- #1
DottZakapa
- 239
- 17
- Homework Statement
- ##\sum_{k=0}^\infty (-1)^\left(k+1\right) \frac {k} {log(k+1)} (2x-1)^k##
- Relevant Equations
- convergence divergence tests
At the exam i had this power series
but couldn't solve it
##\sum_{k=0}^\infty (-1)^\left(k+1\right) \frac {k} {log(k+1)} (2x-1)^k##
i did apply the ratio test (lets put aside for the moment (2x-1)^k ) to the series ##\sum_{k=0}^\infty \frac {k} {log(k+1)}## in order to see to what this limit tends.
##\lim_{n \rightarrow +\infty} \frac {k(1+\frac 1 k)} {log k+log(1+\frac 2 k) } \frac {log k +log (1+\frac 1 k)}{k}=\lim_{n \rightarrow +\infty}\frac {(1+\frac 1 k)} {log k+log(1+\frac 2 k) } \left(log k +log (1+\frac 1 k)\right)##
but from here i don't know how to proceed.
but couldn't solve it
##\sum_{k=0}^\infty (-1)^\left(k+1\right) \frac {k} {log(k+1)} (2x-1)^k##
i did apply the ratio test (lets put aside for the moment (2x-1)^k ) to the series ##\sum_{k=0}^\infty \frac {k} {log(k+1)}## in order to see to what this limit tends.
##\lim_{n \rightarrow +\infty} \frac {k(1+\frac 1 k)} {log k+log(1+\frac 2 k) } \frac {log k +log (1+\frac 1 k)}{k}=\lim_{n \rightarrow +\infty}\frac {(1+\frac 1 k)} {log k+log(1+\frac 2 k) } \left(log k +log (1+\frac 1 k)\right)##
but from here i don't know how to proceed.