Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 10

A convergent series ${\sum }_{n\ge 1}{u}_n$ has partial sums $S_N={\sum }_{n=1}^N{u}_n$ tending to a finite limit $S$. Explicitly, for every $\epsilon >0$ there is an integer $N_0$ such that $|S_N-S|<\epsilon$ whenever $N\ge N_0$. The convergence of the individual terms to zero is necessary but is not sufficient for convergence of the series.

The telescoping series has partial sum $S_N=w_{N+1}-w_1$. Therefore the necessary and sufficient condition is that $w_n$ have a finite limit $L$. The sum is then $L-w_1$.

The comparison test for series states that if $0\le u_n\le v_n$ for all sufficiently large $n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one. Finitely many initial terms do not affect either convergence conclusion.

Subtract the right side from the left and write $g(x)=1-(1+x{)}^{-p}-px(1+x{)}^{-p-1}$. Then $g(0)=0$ and $g^{\prime }(x)=p(p+1)x(1+x{)}^{-p-2}>0$ for $x>0$, so the strict inequality follows. Set $x=1/n$ and multiply by $n^{-p}>0$ to obtainThis decreasing-power difference supplies a telescoping bound for a p-series.

Put $p=k-1>0$. The preceding inequality bounds $n^{-k}<[(n-1{)}^{-p}-n^{-p}]/p$ for $n\ge 2$. For $N\ge 2$, summing the telescoping series givesThese nonnegative partial sums increase and are bounded, so the monotone convergence theorem for sequences proves convergence for every $k>1$. This is the telescoping bound for a p-series, an alternative to the integral test.