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CHAPTER 37
37.101 Determine the nth term of and test for convergence the series
Convergence follows by the alternating series test. However, applying the
limit comparison test with E 1/n we see that
Therefore,
diverges, and the given series is conditionally convergent.
37.102 Determine the nth term of and test for convergence the series
<* n = (-l)"
+1 /(n!)
3 . Use the ratio test
Hence,
the series is absolutely convergent.
37.103 Show by example that the sum of two divergent series can be convergent.
One trivial example is E 1/n + E (-1/n) = 0. Another example is E 1/n + E (1 - n)/n
2 = E 1/n
2
. Of
course, the sum of two divergent series of nonnegative terms must be divergent.
37.104 Show how to rearrange the terms of the conditionally convergent series
series whose sum is 1.
so as to obtain a
Use the first n l positive terms until the sum is >1. Then use the first n 2 negative terms until the sum becomes
<1. Then repeat with more positive terms until the sum becomes >1, then more negative terms until the sum
becomes <1, etc. Since the difference between the partial sums and 1 is less than the last term used, the new
series 1+5 — 5 + 5 — j + 7 + 5 — ••• converges to 1. (Note that the series of positive terms 1 + j + 5 +
• • • and the series of negative terms 2 + 5 + g + • • • are both divergent, so the described procedure always
can be carried out.)
37.105 Test
for convergence.
Use the root test.
(We know that
by Problem 36.15.) Hence, the series converges.
37.106 Show that the root test gives no information when
Let «„ = 1/n. E 1/n is divergent and
On the other hand, let a n = 1/n
2 .
Then E 1/n
2 is convergent and
37.107 Show that the ratio test gives no information when lim \a n + 1 /a n \ = 1.
Let
a n = 1 In.
Then E 1/n is divergent, but
On the other hand, let
a n = 1 In
2 .
Then S 1/n
2 converges, but
37.108 Determine whether
converges.
Use the limit comparison test with E 1/n
3 '
2 .
verges, so does the given series.
(We have used L'Hopital's rule twice.) Since E 1/n
3 '
2 con37.109 Determine whether
converges.
CHAPTER 37
37.101 Determine the nth term of and test for convergence the series
Convergence follows by the alternating series test. However, applying the
limit comparison test with E 1/n we see that
Therefore,
diverges, and the given series is conditionally convergent.
37.102 Determine the nth term of and test for convergence the series
<* n = (-l)"
+1 /(n!)
3 . Use the ratio test
Hence,
the series is absolutely convergent.
37.103 Show by example that the sum of two divergent series can be convergent.
One trivial example is E 1/n + E (-1/n) = 0. Another example is E 1/n + E (1 - n)/n
2 = E 1/n
2
. Of
course, the sum of two divergent series of nonnegative terms must be divergent.
37.104 Show how to rearrange the terms of the conditionally convergent series
series whose sum is 1.
so as to obtain a
Use the first n l positive terms until the sum is >1. Then use the first n 2 negative terms until the sum becomes
<1. Then repeat with more positive terms until the sum becomes >1, then more negative terms until the sum
becomes <1, etc. Since the difference between the partial sums and 1 is less than the last term used, the new
series 1+5 — 5 + 5 — j + 7 + 5 — ••• converges to 1. (Note that the series of positive terms 1 + j + 5 +
• • • and the series of negative terms 2 + 5 + g + • • • are both divergent, so the described procedure always
can be carried out.)
37.105 Test
for convergence.
Use the root test.
(We know that
by Problem 36.15.) Hence, the series converges.
37.106 Show that the root test gives no information when
Let «„ = 1/n. E 1/n is divergent and
On the other hand, let a n = 1/n
2 .
Then E 1/n
2 is convergent and
37.107 Show that the ratio test gives no information when lim \a n + 1 /a n \ = 1.
Let
a n = 1 In.
Then E 1/n is divergent, but
On the other hand, let
a n = 1 In
2 .
Then S 1/n
2 converges, but
37.108 Determine whether
converges.
Use the limit comparison test with E 1/n
3 '
2 .
verges, so does the given series.
(We have used L'Hopital's rule twice.) Since E 1/n
3 '
2 con37.109 Determine whether
converges.