MATH 101 V2A
March 13th – Practice & Homework problems
Practice Problems
1. Make a list of all of the series convergence tests. For each one, come up with one series for which
you can conclude the convergence/divergence of the series by using that test, and one series for which
you can not. Then write down one sentence describing a characteristic of series which are “obvious”
candidates for that test.
Hint: Look in your notes – you shouldn’t need a solution for this!
2. Determine the convergence of the following:
∞
X
(−1)n
√
.
(a)
n−1
n=2
Hint: Use the Alternating Series Test to show that the series converges. Use the Comparison Test
and the p -Test to show that it does not converge absolutely, and therefore converges conditionally.
(b)
∞
X
sin(n) cos(2n)
.
2n
n=1
Hint: Use the Comparison Test and the geometric series
∞
X
1
to show that this series converges
n
2
n=1
absolutely.
(c)
∞
X
(n!)n
.
n5n
n=1
Hint: Use the Comparison Test and the p-series
(d)
∞
X
n=1
n sin
∞
X
1
to show that the series converges absolutely.
5
n
n=1
1
.
n
Hint: Show that limn→∞ n sin( n1 ) = 1 and use the Simple Divergence Test to show that this series
diverges.
3. In class, we constructed an argument to show that
to prove the Alternating Series Test: If
∞
X
∞
X
(−1)n−1
n=1
(−1)
n−1
1
converges. Generalize this method
n
an is an alternating series and if
n=1
(i) lim an = 0
n→∞
(ii) an ≥ an+1 for all n,
then the alternating series converges.
Solution: Let sN denote the N th partial sum of the series. Then
s2N +1 = (a1 − a2 ) + (a3 − a4 ) + . . . + (a2N −1 − a2N ) + a2N +1 ) > a2N +1 > 0,
so the sequence {s2N +1 } is bounded below. Since s2N +3 − s2N +1 = −a2N +2 + a2N +3 < 0, we see that
{s2N +1 } is also monotone decreasing and so, by the MCT, it converges to a number, say A.
Now, we also have that
s2N = a1 + (−a2 + a3 ) + . . . + (−a2N −2 + a2N −1 ) − a2N < a1 ,
so the sequence {s2N } is bounded above. Since s2N +2 − s2N = a2N +1 − a2N +2 > 0, we see that {s2N }
is monotone increasing and so, by the MCT, it converges to a number, say B.
Since A − B = limN →∞ s2N +1 − s2N = limN →∞ a2N +1 = 0, we see that A = B. It now follows that
sequence of partial sums {sN } converges to A, and therefore the series converges.
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