Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 1 / 9F (Analysis)

For $n\ge 2$, $0<n^{-2}\le [n(n-1){]}^{-1}=1/(n-1)-1/n$. The upper bound has telescoping partial sums, bounded by one. The positive partial sums of the original series are consequently bounded by two and increasing, so they converge as a monotone bounded sequence. The series converges absolutely. This also supplies the $p=2$ case of a p-series without requiring its exact sum.

For odd $n$, the denominator is $(n+1{)}^2$; for even $n$, it is $(n-1{)}^2$. There is no zero denominator in the printed index range. For all $n\ge 2$,The comparison test for series with the convergent p-series gives absolute convergence. In fact the consecutive pair $n=2k-1,2k$ contributes $1/(2k{)}^2+1/(2k-1{)}^2$, so the sum is the same as ${\sum }_{m=1}^{\infty }{m}^{-2}$. Positivity justifies this paired reordering; an alternating-sign appearance in the denominator does not make this an alternating series.

Let $a_n$ denote the summand. Its denominator is positive for every $n\ge 1$: odd $n$ gives $n^4+n^2$, while even $n\ge 2$ gives $n^4-n^2$. Dividing numerator and denominator by their highest powers givesThe terms are positive for sufficiently large $n$, so the limit comparison test with the divergent harmonic series gives divergence to $+\infty$. More explicitly, for $n\ge 16$ the numerator is at least $n^3/2$ and the denominator is at most $2n^4$, giving $a_n\ge 1/(4n)$. Any finitely many early negative terms change partial sums by only a finite constant and cannot prevent divergence of this positive tail.

The PDF has the doubly exponential denominator $e^{e^n}$. For its positive summands, the successive-term ratio isThus the ratio test proves absolute convergence. Concretely, the tail is eventually dominated by a geometric series of ratio $1/2$; the polynomial numerator cannot offset the doubly exponential decay.