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### Series Convergence

```
Date: 01/27/2001 at 12:39:50
From: Chris
Subject: Series Convergence

My calculus teacher gave me a review sheet with sample test questions.
The questions on series convergences are really tough!  If you can
help me out with any of them, that would be a great help.

Test each of the following series for convergence. If the series is
alternating, tell whether the convergence is conditional or absolute:

a) infinity
E		((-1)^(k+2))   ((3k + 2) / ((2k + 3)^2))
k = 1

Clarification: -1 is raised to the power k+2; 3k + 2 is divided by
(2k + 3)^2

b) infinity
E		((-1)^(k))	      (2k)!             (e/2)^k
(2^(3k)) (k!)^2
k = 0

Clarification: -1 is raised to the power k; 2k factorial is divided by
2 raised to the power 3k, times k factorial squared; this is all
multiplied by e divided by 2, raised to the power k.

c) infinity
E		((-1)^k)        ((1/square root of k)    -
(1/square root of k + 5))
k = 1

Clarification: -1 is raised to the k; k and k+5 are under square
root signs, and both are in denominator spots.
```

```
Date: 01/28/2001 at 11:29:06
From: Doctor Fenton
Subject: Re: Series Convergence

Hi Chris,

Thanks for writing to Dr. Math.

When I examine series like these, I like to simplify the terms to a
point where I can "see" what is happening. In your first problem,
although it is an alternating series, you should first ignore the
+/- 1 produced by the power of (-1) and check absolute convergence.
The key idea is that for very large k, only the largest power of k in
each term will be significant: 3k+2 is essentially 3k, and (2k+3)^2 is
essentially (2k)^2 = 4k^2.  For very large k, then, the terms are
essentially

3k      3
------ = ---- .
4k^2     4k

If you want to be more formal, write

3k+2         3k (1+[2/(3k)])
--------- = -----------------------
(2k+3)^2    (2k)^2 * (1+[3/(2k)])^2

The second factor in the numerator and denominator is clearly -> 1
as k->oo .

So, essentially, the first problem is like a p-series with p = 1 (the
p-series is the sum of the reciprocal p-th powers of the positive
integers), so this is like the harmonic series (or better, 3/4 times
the harmonic series), and you should expect absolute divergence and
conditional convergence of the alternating series. If you have the
Limit Comparison Test, you can use this; otherwise you need to make
a direct comparison with, say, (1/2) times the harmonic series to show
absolute divergence.

To apply the Alternating Series Test, you need to show that the terms
alternate in sign (given), and that the terms decrease to 0. The
expression above shows that the limit is 0, but you need to show that
the formula for the terms is decreasing. Try showing that f'(x)<0 for

f(x)=(3k+2)/(2k+3)^2 .

The second series, with factorials and exponentials, is tailor-made
for the Ratio Test. If that doesn't work, you've got a really
difficult series (but in your case, it WILL work, and it gives you the

For the third problem, it's difficult to see what's happening because
the terms consist of small numbers being subtracted. Rewrite them as

1           1         sqrt(k+5) - sqrt(k)
------- - ---------- =  ---------------------
sqrt(k)    sqrt(k+5)      sqrt( k*(k+5))

but that's still unclear, since there's still a subtraction in the
numerator, so rationalize the numerator:

sqrt(k+5) - sqrt(k)   sqrt(k+5) + sqrt(k)
-------------------- * -------------------
sqrt( k*(k+5))     sqrt(k+5) + sqrt(k)

(k+5) - k
= --------------------------------------
sqrt( k*(k+5)) [sqrt(k+5) + sqrt(k)]

5
= --------------------------------------
sqrt( k*(k+5)) [sqrt(k+5) + sqrt(k)]

Again, for large k, the first term is sqrt(k(k+5)) ~ k, and the second
term is a sum of two terms, each of which is like sqrt(k), so the
series is like

5
--------
k^(3/2)

which is another p-series, whose behavior you should know.

If you have further questions, please write again.

- Doctor Fenton, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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