#### To determine

**To show:** The
eSn>n+1
and the series is diverges.

#### Explanation

**Proof:**

Let
Sn
is the
nth
partial sum of the harmonic series
Sn=∑n=1∞1n
.

Consider the series
eSn
as shown below,

eSn=e(1+12+13+14+⋯+1n)=e1⋅e12⋅e13⋅e14⋯e1n

Since
en>1+n
, the series can be written as shown below,

e1⋅e12⋅e13⋅e14⋯e1n>(1+1)⋅(1+12)⋅(1+13)⋯(1+1n)eSn>(21)⋅(32)⋅(43)⋯(nn−1)(n+1n)eSn>n+1

Apply the natural logarithm on both sides as shown below,

ln(eSn)>ln(n+1)Sn>ln(n+1)

Apply the limit on both sides as shown below,

limn→∞Sn>limn→∞ln(n+1)limn→∞Sn>∞

Thus,
{sn}
is increasing and it sum greater then infinity.

That is,
limn→∞Sn=∞
.

Therefore, the harmonic series is diverges.