Question

We have seen that the harmonic series is a divergent series whose terms approach 0. Show that \(\sum\limits_{n = 1}^\infty {\ln (1 + \frac{1}{n})} \) is another series with this property.

Short answer

By expansion and logarithmic properties, it can be shown that the given series is divergent whose terms approach 0, similar to the harmonic series.

Step-by-step solution

 

A series can be divergent even if general term isn’t\(0\)

The series \(\sum\limits_{n = 1}^\infty {\ln (1 + \frac{1}{n})} \)has the general term \({a_n} = \ln (1 + \frac{1}{n})\).So as \(n \to \infty \), \({a_n} \to \ln 1 \to 0\)..

but the series is divergent, as is shown below.

\(\ln (1 + \frac{1}{n}) = \ln \frac{{n + 1}}{n} = \ln (n + 1) - \ln (n)\) ().

 

Expansion of series

Expanding by putting values of n, starting from n=1, we obtain:

\(\begin{aligned}{c}(\ln 2 - \ln 1) + \left( {\ln 3 - \ln 2} \right) + \left( {\ln 4 - \ln 3} \right) + ......... + \left( {\ln \left( {n + 1} \right) - \ln n} \right) = \ln \left( {n + 1} \right) - \ln n\\ = \ln \left( {n + 1} \right)\end{aligned}\)

where the first term from each parenthesis cancels out with the second term from the next parenthesis.

Therefore,

S=\(\mathop {\lim }\limits_{n \to \infty } \ln (n + 1) = \infty \)

So, we see that though the general term tends to 0, the series is divergent.