I'm having trouble verifying my appoarch to the problem in $(1)$, much of efforts can be seen in the sections titled $\text{Lemma}$, I'm specifically stuck where $(1.6)$ would the absolute convergence of the double series imply anything about the absolute convergence of the product in $(1)$ ?


Prove that

$$\prod_{n=1}^{\infty}\bigg\{(1-\frac{z}{n})^{n^{k}}\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) \bigg\}$$

where $k$ is any positive integer, converges absolutely for all values of $z.$

$\text{Lemma (1.2)}$

On the RHS side of $(1)$, one can note the following in $(1.3)$


$$\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) = \prod_{m=1}^{k+1}e^{\frac{n^{k-m}z^{m}}{m}} \text{.}$$


The developments in $(1.3)$ is due to the notion that one can go from sum to product, where:

$$\exp \sum s_n = \prod e^{s_n}$$

Using what we've discovered in $(1.3)$, we can make the following deductions in $(1.5)$


$$\log \prod_{n=1}^{\infty}\bigg\{(1-\frac{z}{n})^{n^{k}}\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) \bigg\} = \sum_{n=1}^{\infty}\log \bigg\{(1-\frac{z}{n})^{n^{k}}\prod_{m=1}^{k+1}e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}.$$

$\text{Lemma (1.3)}$

Distributing the $\log$, on the RHS side of $(1.5)$, we obtain a product of two series in $(1.6)$


$$\sum_{n=1}^{\infty}\log \bigg\{(1-\frac{z}{n})^{n^{k}}\prod_{m=1}^{k+1}e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}=\bigg\{\sum_{n=1}^{\infty} \log (1-\frac{z}{n})^{n^{k}} + \sum_{m=1}^{k+1}\log e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}.$$

Finally, condensing our coefficients we have in $(1.7)$


$$\sum_{n=1}^{\infty}n^{k}\log \big(1-\frac{z}{n} \big) + \sum_{m=1}^{k+1} \frac{n^{k-m}z^{m}}{m} \log(e).$$

In summary,

$$\log\left(\prod_{n=1}^{\infty}\bigg\{(1-\frac{z}{n})^{n^{k}}\exp \bigg( \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg) \bigg\}\right)=$$ $$\bigg\{ \sum_{n=1}^{\infty}n^{k}\log \big(1-\frac{z}{n}\big)\bigg\} + \bigg\{\sum_{m=1}^{k+1} \frac{n^{k-m}z^{m}}{m} \log(e).\bigg\}$$

  • $\begingroup$ Which books is this taken from? $\endgroup$
    – John Mayne
    Dec 23, 2017 at 22:15
  • $\begingroup$ @Zophikel Corrected guessed misprints. Please check. $\endgroup$ Dec 26, 2017 at 3:10
  • $\begingroup$ Is $z$ real? What does it mean for an infinite product to converge absolutely? $\endgroup$
    – zhw.
    Dec 26, 2017 at 7:45
  • $\begingroup$ It seems you would be unaware that $$\log e=1$$ Is that so? $\endgroup$
    – Did
    Dec 26, 2017 at 9:32
  • $\begingroup$ @Did This is not so clear. :-) According to an article “Common Logarithm” of CRC Concise Encyclopedia of Mathematics “The logarithm in base $10$. The notation $\log x$ is used by physicists, engineers, and calculator keypads to denote the common logarithm. However, mathematicians generally use the same symbol to mean the natural logarithm ln, $\ln x$. Worse still, in Russian literature the notation $\lg x$ is used to denote a base-$10$ logarithm, which conflicts with the use of the symbol lg to indicate the logarithm to base 2. $\endgroup$ Dec 26, 2017 at 10:32

2 Answers 2



$$\sum_{n=1}^{\infty}\log \bigg\{(1-\frac{z}{n})^{n^{k}}\prod_{m=1}^{k+1}e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}=\bigg\{\sum_{n=1}^{\infty} \log (1-\frac{z}{n})^{n^{k}} + \sum_{m=1}^{k+1}\log e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}.$$

I think this is wrong and the right hand side should be

$$R=\sum_{n=1}^{\infty}\bigg\{ \log (1-\frac{z}{n})^{n^{k}} + \sum_{m=1}^{k+1}\log e^{\frac{n^{k-m}z^{m}}{m}} \bigg\}.$$

From now I suppose plausible claims that $\log$ means $\log_e=\ln$ and $z<1$ in order to take $\log$’s. Then the last expression can be further reduced to $$\sum_{n=1}^{\infty}\bigg\{ n^{k}\log (1-\frac{z}{n}) + \sum_{m=1}^{k+1}\frac{n^{k-m}z^{m}}{m} \bigg\}.$$

PS. I am sorry, I was suddenly distracted by an UFO. :-) I’ll finish my answer later.

OK, since I was not abducted, I can continue.

A natural way to prove the absolute convergence of the series for $R$ is the following. By Taylor's formula in Lagrange form for a function $\log (1+x)$, for each $n$ there exists a number $c_n$ between $0$ and $-\frac zn$ such that

$$\log\left(1-\frac zn\right)=-\sum_{m=1}^{k+1}\frac{z^{m}n^{-m}}{m}+r_n,\mbox{where }r_n=-\frac 1{k+2}\left(\frac {z}{(1+c_n)n}\right)^{k+2}.$$

So $$R=\sum_{n=1}^{\infty} n^kr_n.$$

Then for $z\le 0$ we have $$| n^kr_n |=\left|\frac 1{(k+2)n^2}\left(\frac {z}{1+c_n}\right)^{k+2}\right|\le \left|\frac {1}{(k+2) n^2}\cdot z^{k+2}\right|.$$

Since the series $\sum_{n=1}^{\infty}\frac {1}{n^2}$ converges (to $\frac{\pi^2}6$), the series for $R$ converges absolutely for each fixed $z\le 0$.

For $0<z<1$ we have $$| n^kr_n |=\frac 1{(k+2)n^2}\left(\frac {z}{1+c_n}\right)^{k+2}<\frac {z^{k+2}}{(k+2)\left(1-\frac zn\right)^{k+2}n^2}=\frac {1}{(k+2) n^2}\cdot \left(\frac 1z -\frac 1n\right)^{-k-2} .$$

Since the sequence $\left\{\frac 1z -\frac 1n\right\}$ converges to $\frac 1z$, it is bounded, so the series for $R$ converges absolutely for each fixed $0<z<1$.

Remark that for natural $z$ product (1) is zero, because it contains a multiplier $\left(1-\frac{z}{n}\right)^{n^{k}}$, so it diverges.

At last, I remark that if $z>0$ and in product (1) we start to take $\log$’s of multipliers with $n>z$ then we may drop the condition $z<1$, and similarly to the above conclude that the series for $R$ converges absolutely for each fixed $z\in\Bbb R\setminus\Bbb N$.

  • $\begingroup$ ahh okay what other ways could show that the series converges ? $\endgroup$
    – Zophikel
    Dec 27, 2017 at 0:06
  • $\begingroup$ @Zophikel I think it is hard to estimate the members of the series for $R$ otherwise, because, I guess, we need to collapse their long expression for that. Also, if we wish to consider complex $z$, we may encounter additional problems, both technical and principal. $\endgroup$ Dec 27, 2017 at 6:28

By definition, the infinite product $\prod_{n=1}^\infty (1+a_n)$ converges absolutely iff $\prod_{n=1}^\infty (1+|a_n|)<\infty.$ This happens iff $\sum_{n=1}^{\infty}|a_n|<\infty.$ I'll show your infinite product converges absolutely for all $z\in \mathbb C.$


$$s_n(z) = n^k\left (\frac{z}{n} + \frac{1}{2}\left (\frac{z}{n}\right)^2 + \frac{1}{3}\left (\frac{z}{n}\right)^3 + \cdots + \frac{1}{k+1}\left (\frac{z}{n}\right)^{k+1}\right) .$$

Your product is then

$$\prod_{n=1}^{\infty}(1-z/n)^{n^k}\exp (s_n(z)).$$

Fix $z\in \mathbb C.$ We want to show

$$\tag 1 \sum_{n=1}^{\infty}|(1-z/n)^{n^k}\exp (s_n(z)) - 1| <\infty.$$

Now for $u\in \mathbb C,|u|<1,$ $\log (1-u) = -(u+u^2/2 + u^3/3 + \cdots).$ We can write this as

$$\log(1-u) = -(u+u^2/2 + \cdots + u^{k+1}/(k+1) + r(u)).$$

Let's note that if $|u| \le 1/2,$ then

$$\tag 2 |r(u)| \le \sum_{m=k+2}^\infty |u|^m/m \le \sum_{m=k+2}^\infty |u|^m = |u|^{k+2}(1/(1-|u|) \le 2|u|^{k+2}.$$

For $n$ large,

$$ (1-z/n)^{n^k} = \exp(n^k \log (1-z/n)) = \exp(-s_n(z) + n^kr(z/n)).$$

For such $n,$ the $n$th term in $(1)$ is

$$\tag 3 |\exp(-s_n(z) + n^kr(z/n))\exp (s_n(z)) - 1| = |\exp(n^kr(z/n))-1|.$$

Now for some constant $C,$ $|e^u-1| \le C|u|$ for $|u|<1.$ Thus using $(2),$ and again for large $n,$ $(3)$ is bounded above by

$$ C|n^kr(z/n)| = C|n^k2(z/n)^{k+2}|= 2C|z|^{k+2}\frac{1}{n^2}.$$

Since $\sum_n 1/n^2<\infty,$ we have proved $(1)$ as desired.

  • $\begingroup$ For the convergence of the series in $(1)$ $\tag 1 \sum_{n=1}^{\infty}|(1-z/n)^{n^k}\exp (s_n(z)) - 1|$ is their an alternative way besides using Big O notation $\endgroup$
    – Zophikel
    Jan 1, 2018 at 21:14
  • $\begingroup$ @Zophikel See my edited answer, where I included a few more details. $\endgroup$
    – zhw.
    Jan 2, 2018 at 3:21

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