How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #ln(1/2)+ln(2/3)+ln(3/4)+...+ln(n/(n+1))+...#?

Question

Christopher Allers · Accepted Answer

undefined To find the nth partial sum of the series $ \ln\left(\frac{n}{n+1}\right) $, we need to sum the terms from $ n = 1 $ to $ n = N $.

The nth term is $ \ln\left(\frac{n}{n+1}\right) $.

The nth partial sum, denoted as $ S_N $, is obtained by summing the first $ N $ terms of the series.

To determine whether the series converges, we can investigate its behavior as $ N $ approaches infinity. If the series approaches a finite value as $ N $ goes to infinity, then it converges. Otherwise, it diverges.

To find the sum when it exists, we take the limit of the nth partial sum as $ N $ approaches infinity.

$$ S = \lim_{N \to \infty} S_N $$

$$ S = \lim_{N \to \infty} \sum_{n=1}^{N} \ln\left(\frac{n}{n+1}\right) $$

$$ S = \lim_{N \to \infty} \ln\left(\frac{1}{2}\right) + \ln\left(\frac{2}{3}\right) + \ln\left(\frac{3}{4}\right) + \ldots + \ln\left(\frac{N}{N+1}\right) $$

$$ S = \lim_{N \to \infty} \ln\left(\frac{1}{N+1}\right) $$

$$ S = \ln\left(\lim_{N \to \infty} \frac{1}{N+1}\right) $$

$$ S = \ln(0) $$

As $ N $ approaches infinity, the term $ \frac{1}{N+1} $ approaches 0, which makes the natural logarithm of 0 undefined.

Thus, the series diverges.

Samantha Adkison · Answer

We know that #ln(a/b)=ln(a)-ln(b)#. So, we can rewrite the sum
#ln(1/2)+ln(2/3)+ln(3/4)+...+ln((n-1)/n)+ln(n/(n+1))#
to
#=(ln(1)-ln(2))+(ln(2)-ln(3))+(ln(3)-ln(4))+…+(ln(n-1)-ln(n))+(ln(n)-ln(n+1))#. Slightly rearrange the terms:
#=ln(1)+(ln(2)-ln(2))+(ln(3)-ln(3))+…+(ln(n-1)-ln(n-1))-ln(n+1)# See how almost all of the terms cancel out? This is called a telescoping sum. Everything simplifies to
#=ln(1)-ln(n+1)#
#=ln(1/(n+1))# Now, to find the sum of the infinite series, set #n->oo#:
#lim_(n->oo)ln(1/(n+1))#
#=-oo# So, the sum of the infinite series diverges.