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#### Finding the partial sum, Sn?

33 viewed last edited 1 year ago
Timothy Lee
1
Find the partial sum ,Sn, of the Summation of n=1 to infinite: ln((n+1)/n). Sn = ??? I know the series diverges, but why doesn't it diverge to infinite?
Vivekanand Vellanki
0
How do you know that this series diverges?
Krishna
0
First, let's rewrite "Summation of n=1 to infinite: ln((n+1)/n)" as \lim_{n \to \infty} S_n = \Sigma_{k = 1} ^\infty \ln(\frac{k+1}{k} ). We have to find out if the sequence S_n converges or diverges. Noting that \ln( \frac{k+1} {k}) = \ln{k+1} - \ln{k} we can rewrite S_n = \Sigma_{k = 1} ^n \ln(\frac{k+1}{k} ) as = \ln (2) - \ln (1) + \ln (3) - \ln(2) + \ln(4) -\ln(3)............+\ln(n) - \ln(n-1) + \ln(n+1) - \ln(n) This is a telescoping series where a lot of terms get cancelled out and only two terms remain after cancellation. S_n = -\ln(1)+\ln(n+1) \lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \left( \ln(n+1) - \ln(1) \right) = \infty Therefore, S_n diverges to \infty . Your assertion that it doesn't diverge to \infty is incorrect. In general, a series can still diverge even if does not diverge to \infty . E.g. S_n = (-1)^n is a non-converging series, but it does not diverge to \infty .
Krishna
0
Given Summation of n=1 to infinite: ln((n+1) Modifying the above statment \lim_{n \to \infty} S_n = \Sigma_{n = 1} ^\infty \ln(\frac{n+1}{n}) If the "lim (Sn)" finite: conberges Infinite: diverges Find the partial sum \lim_{n \to \infty} S_n = \Sigma_{n = 1} ^\infty \ln(\frac{n+1}{n}) \lim_{n \to \infty} \ln(\frac{n+1}{n} ) \lim_{n \to \infty} \ln (\frac{n(1+\frac{1}{n}}{n}) \lim_{n \to \infty} \ln(1+\frac{1}{n} ) $\lim _{n \to\infty} [\ln(1)+ln(\frac{1}{n}) )$ 0 + \ln(/\frac{1}{\infty}) \ln(0) \infty The Sn is infinity so it diverges