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$\sum_{n=1}^\infty \log \frac{n+1}{n}$

$\sum_{n=1}^\infty \log \frac{n+1}{n}$ = $\displaystyle\lim_{n \to{+}\infty}(\log2 - \log1)+(\log3-\log2)+...+(\log(n+1)-\log n)$=$\displaystyle\lim_{n \to{+}\infty}\log(n+1)\to \infty$. So, the series diverges.

Is my procedure correct?

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2  
Neat: Since $\log\frac{n+1}{n}<\frac{1}{n}$, this gives a proof that the harmonic series diverges. –  Jonas Meyer May 22 '12 at 5:52
3  
Your answer is correct and you method is also almost right. You need to write it out properly though. Let $\displaystyle S_N = \sum_{n=1}^{N} \log \left( \dfrac{n+1}{n}\right)$. Then we get that $\displaystyle S_N = \sum_{n=1}^{N} \left( \log (n+1) - \log(n) \right)\\ = (\log(2) - \log(1))+(\log(3) - \log(2))+ \cdots + (\log(N+1) - \log(N)) = \log(N+1).$ $$\sum_{n=1}^{\infty} \log \left( \dfrac{n+1}{n}\right) = \lim_{N \rightarrow \infty} \sum_{n=1}^{N} \log \left( \dfrac{n+1}{n}\right) = \lim_{N \rightarrow \infty} S_N = \lim_{N \rightarrow \infty} \log(N+1) = \infty$$ –  user17762 May 22 '12 at 6:04
    
It seems to me that Daniela's and Marvis's solutions are the same except for some pretty minor differences in notation. I agree that the last "$\to$" should be "$=$", and it doesn't hurt to explicitly name the partial sums, or even perhaps use a different symbol to index the partial sum, but I wouldn't say these are needed. That really depends on context, i.e., the audience/grader and what their expectations are for level of detail. –  Jonas Meyer May 22 '12 at 6:19
    
@JonasMeyer True. Thats why I removed my solution and added it as a comment. I wasn't particularly impressed when the $n$ on left side of the equality which ran from $1$ to $\infty$ became $n \rightarrow \infty$ on the right side of the equality though $n$ is just a dummy variable. I also wanted to emphasize the fact that the series should be viewed as the limit of the sequence of partial sums which is what Daniela has also done but it is not completely clear if Daniela understood what she was doing. –  user17762 May 22 '12 at 7:00
    
@Jonas, that is really neat, also note that the premise $\log(1+x) \le x$ is easy to prove when $\log x$ is defined as the inverse of $e^x$ and $e^x$ is defined by its Maclaurin series. –  Dan Brumleve May 22 '12 at 7:43

4 Answers 4

up vote 1 down vote accepted

Your procedure is correct. If you want to write out things more clearly I suggest that you write down the $n$th partial sums $$\begin{align} s_n &= \sum_{i=1}^{n} \log\left(\frac{i+1}{i}\right) \\ &= \sum_{i=1}^{n} \log(i+1) - \log(i) \\ &= [\log(2) - \log(1)] + \dots [\log(n+1) - \log(n)] \\ &= \log(n+1). \end{align} $$ Hence $$ \lim_{n \to \infty} s_n = \lim_{n\to \infty} \log(n+1) = \infty.$$ So then you say that since the limit does not exist, the series is divergent by definition.

Note: The notation is important. It is not correct to write $\lim_{n\to \infty} \log(n+1) \to \infty$, we write $\lim_{n\to \infty} \log(n+1) = \infty.$

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In the comments, @Marvis shows how to clean up your proof. Here's another approach. Let $f_n = \log \frac{n+1}{n}$. Examine the ratio of successive terms for large $n$, $$\frac{f_{n+1}}{f_{n}} = 1 - \frac{1}{n} + O\left(\frac{1}{n^2}\right).$$ Therefore, the series diverges by Gauss's test.

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Your procedure is fine. I think it easiest to remove the logarithm immediately:

$e^{a_k} = \prod_{n=1}^k{\frac{n+1}{n}} = k+1$

So, $a_k = \log{(k+1)} \rightarrow \infty$.

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Your procedure is correct. I suggest another way:

$\log \frac{n+1}{n}$ is positive, and: $\log \left(1+\frac{1}{n}\right)\sim \frac{1}{n}$ for $n \rightarrow\infty$. So, we have:

$\sum \frac{1}{n}$ that diverges.

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