#### To determine

Whether the series is convergent or divergent and obtain the sum if the series is convergent.

#### Answer

The series is convergent to the sum
sin1001−sin100≈−0.336
.

#### Explanation

**Given:**

The series is
∑k=1∞(sin100)k
.

**Result used:**

The geometric series
∑n=1∞arn−1 (or) a+ar+ar2+⋯
is convergent if
|r|<1
and its sum is
a1−r
, where *a* is the first term and *r* is the common ratio of the series.

**Calculation:**

The given series can be written as follows,

∑k=1∞(sin100)k=(sin100)1+(sin100)2+(sin100)3+⋯=(sin100)+(sin100)2+(sin100)3+⋯

Here, the first term of the series is
a=sin100
and the common ratio of the series is,

r=(sin100)2sin100=sin100

The absolute value of *r* is,

|r|=|sin100|=|−0.506|=0.506<1

Since
|r|<1
and by using Result stated above, the series
∑k=1∞(sin100)k
is convergent.

Obtain the sum of the series when
a=sin100 and r=sin100
.

∑k=1∞(sin100)k=sin1001−sin100=−0.5061−(−0.506)=−0.5061.506≈−0.336

Thus, the sum of the series is
sin1001−sin100≈−0.336
.

Therefore, the series is convergent to the sum
sin1001−sin100≈−0.336
.