Series Converge or Diverge?

Question

Determine if the following series converge or diverge:

A. summation of (sin 2 / pi)^n from n=0 to infinity
B. summation of (1/n^(1/3)) from n=1 to infinity
C. summation of (e^n / (e^n +1)) from n=1 to infinity

Answer 1

A. This is a geometric series. Because |sin(2/π)|<1, we know that it converges.
http://en.wikipedia.org/wiki/Geometric_series#Geometric_series

B. This is a power series, but p=1/3. We require p>1 for the series to converge. So it diverges.
http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#.22p-series.22

C. The terms of this series do not approach 0, thus it cannot converge. It diverges.

Answer 2

Hey there!

A. The summation of (sin(2)/pi)^n, where n ranges from 0 to infinity, is a geometric series. Geometric series are any series in the form of ∑a^n, where n ranges from 0 to infinity. The convergence of geometric series states for the common ratio r, in the infinite series ∑r^n

1) If |r|<1, the series converges.
2) If |r|≥1, the series diverges.

The common ratio for the infinite series ∑(sin(2)/pi)^n, where n ranges from 0 to infinity is sin(2)/pi. Since sin(2)/pi is less than 1, the series ∑(sin(2)/pi)^n, converges.

B. The summation of (1/n^(1/3)), where n ranges from 1 to infinity is an example of power series or p-series. The p-series are any series which are in the form ∑1/n^p, where n ranges from 1 to infinity. Note that if n was equal to 0, the series would be in the form ∑1/0 or ∑infinity, or infinity.

The convergence of p-series are mentioned below.

1) If in the series ∑1/n^p, where 0 2) If in the series ∑1/n^p, where p>1, then the series converges.

Since 1/3 is greater than 0, but less than or equal to 1, the series ∑1/n^(1/3), where n ranges from 1 to infinity, diverges.

C. For the summation (e^n/(e^n+1)), where n ranges from 1 to infinity, we need to use the divergence test.

The divergence test stated that limit, as n approaches infinity, of the series A, does not equal to 0, then the series A diverges.

lim (e^n)/(e^n+1) --> Use the divergence test.
n->infinity
lim (e^n)/(e^n) --> Use L'Hopital's rule.
n->infinity
lim 1 --> Cancel out the e^n terms.
n->infinity
1 Evaluate the above limit.

Since 1 does not equal to 0, the series diverges.

Hope it helps!

Answer 3

19geometric series with r= sin2/pi <1 converges
2)1/n^1/3 diverges
3) diverges as lime^n/(e^n+1)=1 and not 0

Answer 4

. They ALL converge, except of B and C. . right?

Answer 5

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