Let $(a_n)_n$ be a sequence of non-negative real numbers.

Prove that $\sum\limits_{n=1}^{\infty}a_n$ converges if and only if the infinite product $\prod\limits_{n=1}^{\infty}(1+a_n)$ converges.

I don't necessarily need a full solution, just a clue on how to proceed would be appreciated.

Could any convergence test like the ratio or the root test help?


4 Answers 4


Hint: Since the $a_n$ are nonnegative, we have the inequalities

$$\sum_{n=1}^N a_n \le \prod_{n = 1}^N (1 + a_n) \le \exp\left\{\sum_{n = 1}^N a_n\right\}\quad (N = 1,2,\ldots)$$

where the second inequality follows from the inequality $1 + x \le e^x$ for all $x \ge 0$.

  • 1
    $\begingroup$ Thanks! I guess one should add "0≤" at the left end of the chain of inequalities because the series must be non-negative in order to rule out divergence to -∞. $\endgroup$
    – Mophotla
    Commented Nov 23, 2016 at 18:51

Hint: Consider $$ \ln\left[\prod_{n=1}^\infty (1+a_n) \right] $$ and note that if $a_n \to 0$, the Taylor series for $\log(1+x)$ is a useful approximation.


A way is to use logarithm and the equivalence $$\ln(1+x) \sim x$$ around the origin.

Interesting to notice is that for complex series (or series not having a constant sign) $\sum z_n$, the absolute convergence of the series implies the convergence of the product $\displaystyle \prod (1+z_n)$, but the converse is not always true. See my counterexamples web site.


Bit of a bump, but here's an approach that doesn't require invoking exp/log:

First, an observation:

Writing "$n\colon N$" as shorthand for "$n\in\left\{0,\dots,N-1\right\}$", remark that for any (not necessarily entrywise nonnegative) sequence $\left(x_{n}\right)_{n\in\mathbb{N}}$, \begin{align*} \left(\prod_{n\colon N} \left(1+x_{n}\right)-\sum_{n\colon N} x_{n}\right)_{N\in\mathbb{N}}\ &=\ \left(1+\sum_{d=2}^{N-1}\sigma_{N, d}\left(\left(x_{n}\right)_{n\colon N}\right)\right)_{N\in\mathbb{N}} \end{align*} with $\sigma_{N,d}$ the $\text{d}^{\text{th}}$ elementary symmetric polynomial in $N$ indeterminates. Trivially (by direct inspection of the coefficients), $$\sigma_{N,d}\left(\left(\left| x_{n}\right|\right)_{n\colon N}\right)\ \geq\ 0$$ and $$\sigma_{N,d}\left(\left(\left| x_{n}\right|\right)_{n\colon N}\right)\ \leq\ \frac{1}{d}\left(\sum_{n\colon N}\left| x_{n}\right|\right)^{d}$$ (insofar as the sum on the right hand side of the latter inequality converges).

Returning to the question, if $\left(\sum_{n\colon N} a_{n}\right)_{N\in\mathbb{N}}$ diverges (i.e., if the sum diverges), then it must grow arbitrarily large (since it is in any case monotonically increasing). But applying the observation above with $\left(x_{n}\right)_{N\in\mathbb{N}}:=\left(a_{n}\right)_{N\in\mathbb{N}}$ gives (using nonnegativity to elide the applications of the absolute value function) $$\left(\prod_{n\colon N} \left(1+a_{n}\right)\right)_{N\in\mathbb{N}}\ \geq\ 1+\left(\sum_{n\colon N} a_{n}\right)_{N\in\mathbb{N}}$$ entrywise in $N$, so $\left(\prod_{n\colon N} \left(1+a_{n}\right)\right)_{N\in\mathbb{N}}$ itself grows arbitrary large and thus diverges, whence one direction of the desired bi-implication.

If, on the other hand, $\left(\sum_{n\colon N} a_{n}\right)_{N\in\mathbb{N}}$ converges (i.e., if the sum converges), then there necessarily exists $L$ such that $\sum_{n\in \mathbb{N}} a_{n+L} < 1$. Applying the observation above with $\left(x_{n}\right)_{N\in\mathbb{N}}:=\left(a_{n+L}\right)_{N\in\mathbb{N}}$ gives (once more using nonnegativity to elide the applications of the absolute value function) \begin{align*} \left(\prod_{n\colon N} \left(1+a_{n}\right)\right)_{N\in\mathbb{N}}\ &=\ \left(\prod_{n\colon L} \left(1+a_{n}\right)\prod_{n\colon N-L} \left(1+a_{n+L}\right)\right)_{N\in\mathbb{N}}\\ &\leq\ \left(\prod_{n\colon L} \left(1+a_{n}\right)\left(1+\sum_{n\colon N-L}a_{n+L}+\sum_{d=2}^{N-L-1}\frac{1}{d}\left(\sum_{n\colon N-L}a_{n+L}\right)^{d}\right)\right)_{N\in\mathbb{N}}\\ &<\ \left(\prod_{n\colon L} \left(1+a_{n}\right)\left(2+\sum_{d=2}^{\infty}\left(\sum_{n\in \mathbb{N}}a_{n+L}\right)^{d}\right)\right)_{N\in\mathbb{N}}\\ &<\ \left(\prod_{n\colon L} \left(1+a_{n}\right)\left(2+\frac{1}{1-\sum_{n\in \mathbb{N}}a_{n+L}}\right)\right)_{N\in\mathbb{N}} \end{align*} entrywise in $N$, the latter term constant. (The boundedness of the sums retroactively implies their convergence by the nonnegativity of the summands.) Thus $\left(\prod_{n\colon N} \left(1+a_{n}\right)\right)_{N\in\mathbb{N}}$ is (uniformly) bounded, and is clearly monotonically increasing, so converges, whence the other half of the claim in the OP. $\blacksquare$

Remark: A suitable modification of the above argument suffices to establish the more general result that the convergence of $\left(\sum_{n\colon N} \left|a_{n}\right|\right)_{N\in\mathbb{N}}$ implies the convergence of $\left(\prod_{n\colon N} \left(1+a_{n}\right)\right)_{N\in\mathbb{N}}$ even without any nonnegativity assumptions on $\left(a_{n}\right)_{n\in\mathbb{N}}$. (In fact, the convergence of the former even implies that the value to which the product converges is nonzero so long as no multiplicand is zero via something like Bonferroni's inequality.)

Indeed, once more choose $L$ such that $\sum_{n\in\mathbb{N}} \left|a_{n+L}\right| < 1$, and deduce that \begin{align*} \left(\left|\prod_{n\colon N} \left(1+a_{n}\right)\right|\right)_{N\in\mathbb{N}}\ &\leq\ \left(\prod_{n\colon N} \left(1+\left|a_{n}\right|\right)\right)_{N\in\mathbb{N}}\\ &\leq\ \text{(same steps)}\\ &<\ \left(\prod_{n\colon L} \left(1+\left|a_{n}\right|\right)\left(2+\frac{1}{1-\sum_{n\in \mathbb{N}}\left|a_{n+L}\right|}\right)\right)_{N\in\mathbb{N}} \end{align*} Now for any $1>\eta>0$, we can moreover choose $L_{\eta}$ such that $\sum_{n\in\mathbb{N}} \left|a_{n+L_{\eta}}\right| < \eta$. Then for $N_{1}\geq N_{0}\geq L_{\eta}$, we have \begin{align*} &\left|\prod_{n\colon N_{1}} \left(1+a_{n}\right) - \prod_{n\colon N_{0}} \left(1+a_{n}\right)\right|\\ &=\ \left|\prod_{n\colon L} \left(1+a_{n}\right)\right|\left|\left(-1+\prod_{n\colon N_{1}-L} \left(1+a_{n+L}\right)\right)-\left(-1+\prod_{n\colon N_{0}-L} \left(1+a_{n+L}\right)\right)\right|\\ &<\ \left|\prod_{n\colon L} \left(1+a_{n}\right)\right|\left(\left|-1+\prod_{n\colon N_{1}-L} \left(1+a_{n+L}\right)\right|+\left|-1+\prod_{n\colon N_{0}-L} \left(1+a_{n+L}\right)\right|\right)\\ &\leq\ 2\left|\prod_{n\colon L} \left(1+a_{n}\right)\right|\left(\eta+\frac{\eta^{2}}{1-\eta}\right)\\ &<\ 2\prod_{n\colon L} \left(1+\left|a_{n}\right|\right)\left(2+\frac{1}{1-\sum_{n\in \mathbb{N}}\left|a_{n+L}\right|}\right)\left(\eta+\frac{\eta^{2}}{1-\eta}\right) \end{align*} (the last inequality by mimicking the argument used for the original value of $L$). As this latter expression clearly converges to $0$ in $\eta$ (being directly proportional to $\eta^{1}$ in the limit), it follows that $\left(\prod_{n\colon N} \left(1+a_{n}\right)\right)_{N\in\mathbb{N}}$ is Cauchy, whence converges, as claimed. $\Box$


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