Series of Geometric Means Converges

Question

I'm trying to prove the following:

Let $\{b_{n}\}_{n \ge 1}$ be a sequence of strictly positive terms such that $$\sum_{n = 1}^{\infty}b_{n}$$ converges. Set $\gamma_{n} = (b_{1} \cdots b_{n})^{1/n}$. Prove $$\sum_{n = 1}^{\infty}\gamma_{n}$$ converges.

Attempt: Letting $b = \sum b_{n}$, and using AM-GM: \begin{align*} \sum_{n = 1}^{\infty}(b_{1} \cdots b_{n})^{\frac{1}{n}} &< \sum_{n = 1}^{\infty}b_{1}\cdots b_{n} \\ &< \sum_{n = 1}^{\infty}\frac{1}{n^{n}}(b_{1}+\cdots+b_{n})^{n} \\ &< \sum_{n = 1}^{\infty}\left(\frac{b}{n}\right)^{n} \end{align*} This last series converges by the root test, so $\sum\gamma_{n}$ converges by comparison, given $b_{1} \cdots b_{n} \ge 1$ for all $n$. If $b_{1} \cdots b_{n} < 1$ for finite $n$, then we can reduce this case to the previous argument by adding a sufficiently large constant to $\sum(\frac{b}{n})^{n}$. My problem is that I dont know how to deal with the case: $b_{1} \cdots b_{n} < 1$ for infinite $n$...

Answer 1

Here is another variation of the theme by also introducing factors to assure the convergence when applying AM-GM.

We obtain \begin{align*} \color{blue}{\sum_{n=1}^\infty}&\color{blue}{\left(b_1b_2\cdots b_n\right)^{1/n}}\\ &=\sum_{n=1}^\infty\frac{\left(b_1c_1b_2c_2\cdots b_nc_n\right)^{1/n}}{\left(c_1c_2\cdots c_n\right)^{1/n}}\tag{1}\\ &\leq \sum_{n=1}^\infty\frac{b_1c_1+b_2c_2+\cdots+ b_nc_n}{n\left(c_1c_2\cdots c_n\right)^{1/n}}\tag{2}\\ &=\sum_{n=1}^\infty b_nc_n\sum_{j=n}^\infty\frac{1}{j\left(c_1c_2\cdots c_j\right)^{1/j}}\tag{3}\\ &=\sum_{n=1}^\infty b_nc_n\sum_{j=n}^\infty \frac{1}{j(j+1)}\tag{4}\\ &=\sum_{n=1}^\infty b_n\frac{c_n}{n}\tag{5}\\ &=\sum_{n=1}^\infty b_n\left(1+\frac{1}{n}\right)^n\tag{6}\\ &\,\,\color{blue}{\leq e\sum_{n=1}^\infty b_n}\tag{7} \end{align*}

and the claim follows.

Comment:

• In (1) we introduce factors $c_j$ which we will appropriately set to assure convergence of the series.

• In (2) we apply the AM-GM.

• In (3) we do some rearrangements.

• In (4) we inductively set $(c_1c_2\cdots c_j)^{1/j}=j+1$ which gives us a nicely telescoping series.

• In (5) we use the telescoping property $\sum_{j=n}^\infty\frac{1}{j(j+1)}=\sum_{j=n}^\infty\left(\frac{1}{j}-\frac{1}{j+1}\right)=\frac{1}{n}$.

• In (6) we find $$c_n=\frac{c_1c_2\cdots c_n}{c_1c_2\cdots c_{n-1}}=\frac{(n+1)^n}{n^{n-1}}=n\left(1+\frac{1}{n}\right)^n$$.

• In (7) we use the bound $1+x< e^x$ with $x=\frac{1}{n}, n=1,2,\ldots$.

Note:

• The inequality (7) is known as Carleman's Inequality.

• This answer follows the solution of Problem 2.4 in The Cauchy-Schwarz Master Class by J. M. Steele.

Answer 2

Using AM-GM,

$$\gamma_n = \left(\prod_{k=1}^n b_k\right)^{1/n} = \left(\frac{1}{n!}\prod_{k=1}^nkb_k\right)^{1/n} \leqslant \frac{1}{n(n!)^{1/n}}\sum_{k=1}^n k b_k$$

Using the inequality $(n!)^{-1/n} \leqslant e/(n+1)$ (which follows from Stirling's approximation) we have

$$\sum_{n=1}^m \gamma_n \leqslant \sum_{n=1}^m\frac{1}{n(n!)^{1/n}}\sum_{k=1}^n k b_k = \sum_{n=1}^m\frac{1}{n(n!)^{1/n}}\sum_{k=1}^m k b_k \mathbf{1}_{k \leqslant n} \\ = \sum_{k=1}^m b_k k\sum_{n=k}^m\frac{1}{n(n!)^{1/n}} \\\leqslant \sum_{k=1}^m b_k k\sum_{n=k}^m\frac{e}{n(n+1)} \\ \leqslant \sum_{k=1}^m b_k k\sum_{n=k}^\infty\frac{e}{n(n+1)} \\= e \sum_{k=1}^mb_k, $$

and convergence of $\sum b_n$ implies convergence of $\sum \gamma_n$.

Answer 3

You can also prove this using Hardy's inequality:

Set $a_n=\sqrt b_n.$ Then we have $\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty a_n^2 <\infty.$ Now write

\begin{align} \sum_{n=1}^\infty \gamma_n &= \sum_{n=1}^\infty ((a_1\cdots a_n)^{1/n})^2 \\ &\le \sum_{n=1}^\infty \left(\frac{1}{n}\sum_{k=1}^n a_k\right)^2 \\ &< 4\sum_{n=1}^\infty a_n^2 < \infty. \end{align}