Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 20

The summand is $\sqrt{n}/\sqrt{n+1}=\sqrt{n/(n+1)}\to 1$, rather than zero. The necessary term test for divergence fails. Therefore the positive-term series diverges to $+\infty$.

For $a_n=n^2/n!$, the ratio test givesThus the series converges absolutely. As a check, $n^2=n(n-1)+n$ and the exponential power series give the sum $e+e=2e$.

For $a_n=n^n/n!$, successive terms satisfyThey eventually grow by a factor strictly greater than one and cannot tend to zero. Hence the series diverges to $+\infty$, by the ratio test or the term test for divergence.

The Maclaurin series give $\sin (3x)=3x+O(x^3)$ and $\sinh x=x+O(x^3)$. Dividing by $x$ in a punctured neighborhood of zero gives

The linear terms cancel: $\tan x=x+x^3/3+O(x^5)$, whereas $\tanh x=x-x^3/3+O(x^5)$. Also $\sinh x-x=x^3/6+O(x^5)$. Their leading nonzero terms therefore giveThe Maclaurin series remainder orders justify division after factoring out $x^3$.

To retain the finite term after cancellation, expand far enough before dividing. The Maclaurin series areThusand hence