# Absolute convergence of $\sum\limits_n\left( \log (1-\frac{z}{n})^{n^{k}} + \sum\limits_{m=1}^{k+1}\log e^{n^{k-m}z^m/m} \right)$?

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 $(1)$ ?

$(1)$

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 $(1.3)$

$(1.3)$

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

$\text{Remark}$

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 $(1.5)$

$(1.5)$

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

$\text{Lemma (1.3)}$

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

$(1.6)$

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

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

$(1.7)$

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

In summary,

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

• Which books is this taken from? Dec 23, 2017 at 22:15
• @Zophikel Corrected guessed misprints. Please check. Dec 26, 2017 at 3:10
• Is $z$ real? What does it mean for an infinite product to converge absolutely?
– zhw.
Dec 26, 2017 at 7:45
• It seems you would be unaware that $$\log e=1$$ Is that so?
– Did
Dec 26, 2017 at 9:32
• @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. Dec 26, 2017 at 10:32

$$(1.6)$$

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

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 $$0 we have $$| 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.

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 $$z\in\Bbb R\setminus\Bbb N$$.

• ahh okay what other ways could show that the series converges ? Dec 27, 2017 at 0:06
• @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. 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 $z\in \mathbb C.$

Define

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

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

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.

• 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 Jan 1, 2018 at 21:14
• @Zophikel See my edited answer, where I included a few more details.
– zhw.
Jan 2, 2018 at 3:21