Determine whether the series is convergent or divergent. If it is convergent, find its sum.
Whether the series is convergent or divergent and obtain the sum if the series is convergent.
The series is divergent.
The series is ∑n=1∞(35n+2n).
(1) The geometric series ∑n=1∞arn−1 (or) a+ar+ar2+⋯ is convergent if |r|<1 and its sum is a1−r.
(2) The geometric series ∑n=1∞arn−1 (or) a+ar+ar2+⋯ is divergent if |r|≥1, where a is the first term and r is the common ratio of the series.
The given series can be written as follows,
Here, ∑n=1∞1n is harmonic series and it is divergent.
Note that, a nonzero multiple of a divergent series is also divergent.
Therefore, the series 2⋅∑n=1∞1n is divergent.
Consider the series, ∑n=1∞(15)n=(15)+(15)2+(15)3+⋯.
Clearly, it is geometric series with the first term of the series, a=15 and the common ratio of the series is,
The absolute value of r is,
Since |r|<1 and by the Result (1) stated above, the series is convergent.
Thus, the series ∑n=1∞(15)n is convergent.
Note that, a nonzero multiple of a convergent series is also convergent.
Therefore, the series 3⋅∑n=1∞(15)n is convergent.
Note that, the sum of a convergent and a divergent series is divergent.
Since the series 3⋅∑n=1∞(15)n is convergent and 2⋅∑n=1∞1n is divergent, then their sum ∑n=1∞(2e)n+∑n=1∞(4e)n is divergent.
From equation (1), the series ∑n=1∞(35n+2n) is divergent.
Therefore, the series is divergent.