Question

\(\sum\limits_{n = 1}^\infty {\left( {{e^{1/n}} - {e^{1/n + 1}}} \right)} \)

Short answer

A convergent series is a series whose partial sums tends to a specific number also called a limit. A divergent series is a series whose partial sums, by contrast don’t approach a limit.

Step-by-step solution

Given that:-

\(\sum\limits_{n = 1}^\infty {\left( {{e^{1/n}} - {e^{1/n + 1}}} \right)} \)

\({S_n} = \sum\limits_{n = i}^\infty {{e^{1/i}} - {e^{1/i + 1}}} \)

\({S_n} = ({e^{1/1}} - {e^{1/2}}) + ({e^{1/2}} - {e^{1/3}}) + \left( {{e^{1/3}} - {e^{1/4}}} \right)\)

\({S_n} = e - {e^{1/n + 1}}\)

Finding its sum:

\(\sum\limits_{n = 1}^\infty {\left( {{e^{1/n}} + {e^{1/n + 1}}} \right) = \mathop {\lim }\limits_{n \to \infty } {S_n}} \)

\(\sum\limits_{n = 1}^\infty {\left( {{e^{1/n}} + {e^{1/n + 1}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left( {e - {e^{1/n + 1}}} \right)} \)

\(\sum\limits_{n = 1}^\infty {\left( {{e^{1/n}} + {e^{1/n + 1}}} \right) = e - 1} \)

Limit exist

Hence, the series is convergent.