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Consider the series

$$\sum_{k = 1}^{\infty} \ln\frac{k+1}{k+2}$$

This is what I tried:

Since this is not a geometric series I tried to compute parital sums. So I did:

$$\sum_{k = 1}^{\infty} \ln({k+1})-ln({k+2})$$

I computed the first couple sums and noiced that as k approached infinity, the sum goes towards negative infinity. So to show this could I say:

$$\lim _{k\to ∞} = \lim _{k\to ∞} \ln(k)-\ln(k+1)$$

mookid
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Overclock
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  • You have done very well, jut write out the first few terms, the first five or so will do, and you will see the pattern. – ReverseFlowControl Nov 13 '14 at 22:38
  • Take the partial sums: start with $\sum_{k=1}^2$, then $\sum_{k=1}^3$, for example, and that should let you figure out $\sum_{k=1}^N$ for any positive integer $N$. Then take $N \to \infty$. – Platehead Nov 13 '14 at 22:39
  • Looks like you've got plenty of answers here. Just so you know, this type of series is called "telescoping" – graydad Nov 13 '14 at 22:44
  • Great I will be sure to look that up and mention it to the class! – Overclock Nov 13 '14 at 22:46

3 Answers3

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$$S_1 = \ln(2) - \ln(3)$$ $$S_2 = \ln(2) - \ln(3) + \ln(3) - \ln(4) = \ln(2) - \ln(4)$$ and more generally, $$S_N = \ln(2) - \ln(N+2)$$

Now take the limit as $N \to \infty$. If this converges, you have your sum. If it diverges, then the series diverges.

Joel
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  • So if I take the limit, I notice it goes towars negative infinity, is that correct? – Overclock Nov 13 '14 at 22:47
  • Yep. That would do it. – Joel Nov 13 '14 at 22:52
  • One more thing if you dont mind, should I edit my question to my final correct answer or should I leave it? – Overclock Nov 13 '14 at 22:53
  • Usually people do not change the question that they asked once it has been stated clearly. If you like any answer here, you should accept it. That would indicate that you have resolved the issue. – Joel Nov 13 '14 at 22:55
1

What you did is fine.

You can also (this is the same idea) $$ \sum_{k=1}^n \log\frac{k+1}{k+2} = \log \prod_{k=1}^n \frac{k+1}{k+2} = \log \frac 2{n+2} $$

A brainless solution:

$$ \log\frac{k+1}{k+2} = \log\left(1 - \frac{1}{k+2}\right)\sim - \frac{1}{k+2}<0 $$ and as $\sum - \frac{1}{k+2} = -\infty$ the series has limit $-\infty$ as well.

mookid
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1

Notice that

$$\ln\frac{k+1}{k+2}=\ln\left(1-\frac1{k+2}\right)\sim_\infty-\frac1k$$ and the harmonic series $\sum\frac1k$ is divergent so your series is divergent by comparison.