Can we show the sum of this series to $n$ terms is $n/(3n-1)$?

Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Show that the sum to \(n\) terms of the series \[\frac{1}{2.1}-\frac{1}{5.2}-\frac{1}{8.5}-\cdots+ \frac{1}{(3r-1)(4-3r)}\cdots\] is \[\frac{n}{3n-1}.\]

Given that for \[-1<x<+1, \quad \log_e(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}- \quad\text{to infinity,}\] deduce the sum to infinity of \[x+\frac{x^5}{5}+\frac{x^7}{7}+\cdots.\]

[Note that the original question had \(\log\) meaning ‘\(\log\) to base \(e\)’.]