CHAPTER 38
Power Series
In Problems 38.1-38.24, find the interval of convergence of the given power series. Use the ratio test, unless otherwise
instructed.
38.1
2 x"/n.
for
and diverges for
When
Therefore, the series converges absolutely
the series is
which converges by the alternating series test. Hence, the series converges
we have the divergent harmonic series E l/n. When
for
38.2
E x"/n
2 .
ly for
Thus, the series converges absoluteand diverges for
When * = 1, we have the convergent p-series E l/n
2 . When
the series converges by the alternating series test. Hence, the power series converges for -1 s x s 1.
*=-!,
38.3
E*"/n!.
Therefore, the series converges for all x.
38.4
E nix"
(except when x = 0). Thus, the series converges only for
x = 0.
38.5
E x"/2".
This is a geometric series with ratio x/2. Hence, we have convergence for |j;/2| divergence for |jc|>2. When x = 2, we have El, which diverges. When x = -2, we have E(-l)",
which is divergent. Hence, the power series converges for -2 < x < 2.
38.6
Ex"/(rt-2").
Thus, we have convergence for
|*| < 2, and divergence for |jd>2. When x = 2, we obtain the divergent harmonic series. When x = —2.
we have the convergent alternating series E (-l)7n. Therefore, the power series converges for — 2sjc<2.
38.7
E nx".
So we have convergence f&r \x\ < 1, and divergence foi
\x\ > 1. When x = 1, the divergent series E n arises. When x = — 1, we have the divergent series
E (— l)"n. Therefore, the series converges for — l 38.8
E 3"x"/n4".
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Thus, we have convergence
for
and divergence for
For
we obtain the divergent series E l/n, and, for
we obtain the convergent alternating series E(-l)"/n. Therefore, the power series converges for