The first series is <telescoping series>[telescoping], since
$$
\frac1{n^2+n}=\frac1n-\frac1{n+1}.
$$
Its $N$th partial sum is $1-1/(N+1)$, so
$$
\boxed{\sum_{n=1}^{\infty}\frac1{n^2+n}=1},
$$
a <rational number>.

The second series converges absolutely, for example by the <ratio test>, and is the odd part of the <exponential series>:
$$
\sum_{n=1}^{\infty}\frac1{(2n-1)!}
=\sum_{k=0}^{\infty}\frac1{(2k+1)!}
=\sinh1=\frac{e-e^{-1}}2.
$$
This value is irrational. Indeed, if $a=\sinh1$ were algebraic, then $e$ would satisfy
$$
e^2-2ae-1=0,
$$
making $e$ algebraic over the algebraic numbers and hence algebraic, contrary to the <Hermite theorem on the transcendence of e>. Thus the second sum is in fact <transcendental number>[transcendental].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 2E / Solution

{c}
{title2=$e\notin\overline{\mathbb Q}$}

The Hermite theorem states that the base $e$ of the natural logarithm is a <transcendental number>.

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 Incoming links: Hermite theorem on the transcendence of e