# Prove that $\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)$ converges and find its sum

Since this is not a geometric series, I know that I should use the definition of a convergent series, so $$S_n = \sum_{i=1}^n \ln\left(\frac{i(i+2)}{(i+1)^2}\right)$$ After this, I tried two different ways: 1) I simplified the fraction to read $$\frac{\ln(i^2+2i)}{i^2+2i+1}$$ and then I used long division to get $$\ln((1)-\frac{1}{(i+1)^2})$$ However, once I start plugging in i starting at 1, I don't know where to go from there. 2) I simplified the equation to read $\ln(i^2+2i)-\ln(i^2+2i+1)$, but again, once I start plugging in i starting at 1, I don't know where to go from there. What can I do?

Note: Originally the series was presented as $\sum_{n=1}^\infty \frac{\ln(n(n+2))}{(n+1)^2}$ which is what some of the answers below are adressing.

• Don't you mean to use $i$'s instead of $n$'s in the first equation? – Bobson Dugnutt Feb 28 '16 at 20:21
• Convergence: $$\frac{\ln(i(i+2))}{(i+1)^2}\leqslant\frac{\ln((i+1)^2)}{(i+1)^2}=\frac{4{}{}{}{}\ln\sqrt{i+1}}{(i+1)^2}\leqslant\frac4{(i+1)^{3/2}}$$ Value of the sum: sure you are asked this? – Did Feb 28 '16 at 20:54
• Are you sure that you take the logarithm of the numerator, and not of the whole fraction? – D. Thomine Feb 28 '16 at 22:17
• $$S_n=\sum_{i=1}^n\left(\ln(i)+\ln(i+2)-2\ln(i+1)\right)=\sum_{i=1}^n\ln(i)+{}{}{}{}{}\sum_{i=3}^{n+2}\ln(i)-2\sum_{i=2}^{n+1}\ln(i)=\ldots$$ – Did Feb 28 '16 at 23:29
• This question is sufficiently different from the original question, which had answers, that it would have been better to ask a new question. – Eric Towers Feb 29 '16 at 13:52

The series $$\sum_{n=1}^{\infty}\ln\left(\frac{n(n+2)}{(n+1)^2}\right)$$ converges, using the telescoping technique: \begin{align} \sum_{n=1}^{\infty}\ln\left(\frac{n(n+2)}{(n+1)^2}\right)&=\left[\ln(1)+\ln(3)-2\ln(2)\right]+\left[\ln(2)+\ln(4)-2\ln(3)\right]+\left[\ln(3)+\ln(5)-2\ln(4)\right]+\cdots \end{align}

Note that the partial sums are \begin{align} S_1&=\ln(1)+\ln(3)-2\ln(2)\\ S_2&=\ln(1)+\ln(3)-2\ln(2)+\ln(2)+\ln(4)-2\ln(3)\\ &=\ln(1)-\ln(2)-\ln(3)+\ln(4)\\ S_3&=S_2+\ln(3)+\ln(5)-2\ln(4)\\ &=\ln(1)-\ln(2)-\ln(4)+\ln(5) \end{align} And in general, $S_n=\ln(1)-\ln(2)-\ln(n+1)+\ln(n+2)=\ln\left(\frac{n+2}{2(n+1)}\right)$ which approaches $\ln(1/2)$.

The series can be turned into telescopic as follow \begin{align} S_m=\sum_{n=1}^{m}\ln\frac{n(n+2)}{(n+1)^2}&=\sum_{n=1}^m(\ln{(n+2)}+\ln n-2\ln{(n+1)}) \\ &=\sum_{n=1}^m((\ln{(n+2)}-\ln(n+1))-(\ln{(n+1)}-\ln n))) \\ &=\ln{(m+2)}-\ln(m+1)-(\ln{(2)}-\ln(1)) \end{align} So $$\lim_{m\to\infty}S_m=\lim_{m\to\infty}\left(\ln\frac{m+2}{m+1}-\ln2\right)=-\ln2$$

\begin{align} \sum_{k=1}^\infty\log\left(\frac{k(k+2)}{(k+1)^2}\right) &=\log\left(\prod_{k=1}^\infty\frac{k(k+2)}{(k+1)^2}\right)\\ &=\lim_{n\to\infty}\log\left(\prod_{k=1}^n\frac{k(k+2)}{(k+1)^2}\right)\\ &=\lim_{n\to\infty}\log\left(\frac{\Gamma(n+1)}{\Gamma(1)}\frac{\Gamma(n+3)}{\Gamma(3)}\frac{\Gamma(2)^2}{\Gamma(n+2)^2}\right)\\ &=-\log(2)+\lim_{n\to\infty}\log\left(\frac{\Gamma(n+1)\Gamma(n+3)}{\Gamma(n+2)^2}\right)\\[6pt] &=-\log(2) \end{align} The last limit is a consequence of Gautschi's Inequality.

Another Approach

Prove by induction that $$\prod_{k=1}^n\frac{k(k+2)}{(k+1)^2}=\frac12\frac{n+2}{n+1}$$

HINT for the convergence: $$\frac{n(n+2)}{(n+1)^2}=1-\frac{1}{(n+1)^2}$$ thus $$\log\left(\frac{n(n+2)}{(n+1)^2}\right)\sim\frac{-1}{(n+1)^2}.$$ HINT for the sum: use telescopic properties of the summand.

(Edit: The answer below refers to the first version of the question posted. OP eventually substantially changed the question. If someone's search for answers brings them to this answer to this question, at least they will have an answer. I note that given the large gap between the original and edited questions, asking a new question would have been more appropriate.)

Note that $\ln(x)$ has an alternating Taylor series when evaluated around $x=1$, so, for instance, \begin{align*} \ln 3 + \frac{8}{3}(x-1) - \frac{8}{9}(x-1)^2 \leq \ln(x^2(x^2+2)) \leq \ln 3 + \frac{8}{3}(x-1) \text{,} \end{align*} or, perhaps more obviously useful, \begin{align*} \ln 3 + \frac{8}{3}(\sqrt{i}-1) - \frac{8}{9}(\sqrt{i}-1)^2 \leq \ln(i(i+2)) \leq \ln 3 + \frac{8}{3}(\sqrt{i}-1) \text{.} \end{align*} (This gives us the inequalities on the interval $(0,2)$. To extend beyond there, we need to take derivatives and notice the directions of the inequalities remain unchanged (this is not automatic, we have some work to do), so the upper bound starts above and only ever grows faster than the logarithm and the lower bound starts below and only ever grows slower than the logarithm. In fact, the lower bound is eventually negative, so these bounds are rather loose.)

That upper bound will get you what you need, because \begin{align*} \frac{\ln(i(i+2))}{(i+1)^2} &\leq \frac{\ln 3 + \frac{8}{3}(\sqrt{i}-1)}{(i+1)^2} \\ &\leq \frac{\ln 3 + \frac{8}{3}(\sqrt{i}-1)}{i^2} \\ &= \frac{\sqrt{i} \left( \frac{8}{3} - \frac{8}{3\sqrt{i}} + \frac{\ln 3}{\sqrt{i}} \right)}{i^2} \\ &\leq \frac{8}{3} \frac{1}{i^{3/2}} \text{,} \end{align*} where we have used $\frac{8}{3} > \ln(3)$ in the last step to simplify the numerator. This gives convergence by comparison with the geometric series.

You're not going to find the value of this sum without some sort of stupendous trick. Alternatively, we could attack it numerically, but you don't indicate that would meet your needs.