#### To determine

**To show**:
∑n=2∞1n1.3<∫1∞1x1.3
by using the graph; Comment the nature of series.

#### Explanation

**Result used**:

The *p-*series
∑n=1∞1np
is convergent if
p>1
and divergent if
p≤1
.

**Proof**:

From the series
∑n=2∞1n1.3
, the sequence is
an=1n1.3
.

The graph of the function
y=1x1.3
is shown below in Figure 1.

From Figure 1, it is observed that the function
y=1x1.3
is decreasing on
[1,∞)
and the area of the rectangles less than the area under the function
y=1x1.3
.

(a2+a3+a4+⋯)≤∫2∞ydx∑n=2∞an≤∫2∞ydx∑n=2∞1n1.3≤∫2∞1x1.3dx

Consider the series
∑n=2∞1n1.3
and value of
p=1.3
.

Here,
p>1
.

From the Result stated above, it can be concluded that the series
∑n=2∞1n1.3
is convergent.