= Solution
If
$$
\sum_{n=0}^{\infty}\frac1{\lambda_n}<\infty,
$$
then
$$
\mathbb ET_\infty=\sum_{n=0}^{\infty}\mathbb EX_n
=\sum_{n=0}^{\infty}\frac1{\lambda_n}<\infty.
$$
A nonnegative random variable with finite expectation is finite almost surely, so the process explodes.
Conversely, suppose the series diverges. For $s>0$, independence and the <Laplace transform> of an exponential variable give
$$
\mathbb E e^{-sT_\infty}
=\prod_{n=0}^{\infty}\frac{\lambda_n}{\lambda_n+s}
=\exp\left[-\sum_{n=0}^{\infty}
\log\left(1+\frac{s}{\lambda_n}\right)\right].
$$
The sum in the exponent diverges. If infinitely many $\lambda_n\le s$, infinitely many terms are at least $\log2$; otherwise, eventually $s/\lambda_n\le1$ and
$$
\log(1+s/\lambda_n)\ge\frac{s}{2\lambda_n}.
$$
Thus $\mathbb Ee^{-sT_\infty}=0$. Since $e^{-sT_\infty}$ is strictly positive on $\{T_\infty<\infty\}$, this forces $T_\infty=\infty$ almost surely. Hence the <nonexplosion criterion for a pure birth process> is
$$
\boxed{N\text{ is nonexplosive}
\iff\sum_{n=0}^{\infty}\lambda_n^{-1}=\infty}.
$$
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