Expert: Paul Klarreich - 11/24/2005

Question
 Mr. Klarriech,
I never received an answer to the questions below. There are two questions in the paragraph below that I wanted you to answer. Maybe something got mixed up because I never got an answer to those two. Thank you for your help.
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Followup To
Question -
 Mr. Klarreich,
Thank you so much for your advice on the last 3 problems I sent you. It was a huge help to me! Now I have another question. I am supposed to use the Direct Comparison Test to determine if a series converges or diverges. The series is: the summation from n=1 to infinity of
[ 3/{7n+10n^(1/2)}]. I've tried it 5 different ways and none of them work out. Also, is this statement true or false: "If a sub n and f(n) satisfy the requirements of the Integral Test, and if the integral from 1 to infinity of f(x)dx converges, then the summation from n=1 to infinity of a sub n = the integral from 1 to infinity of f(x)dx." I think that it's false, but i'm not sure that I fully understand what it's saying so I thought I would ask. Thank you for your help.
Answer -
Hi, Leslie,

Found them.  They were the ones on (2n+2)! and geometric series?  

Answer
Hi, Leslie,

[THIS IS WHAT I THOUGHT I SENT THE FIRST TIME.  IF YOU DIDN'T GET IT, I CAN UNDERSTAND THAT YOU WERE CONFUSED.]

------------ Your question -----------
Thank you so much for your advice on the last 3 problems I sent you. It was a huge help to me! Now I have another question. I am supposed to use the Direct Comparison Test to determine if a series converges or diverges. The series is: the summation from n=1 to infinity of [ 3/{7n+10n^(1/2)}]
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I wish I could remember which ones they were.  Unfortunately, although this web site keeps all the previous questions, they don't get listed by name, only by subject.  So I have to dig through to see if I gave you good answers or not.  I'll take your word for it.
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That's the sum of:
     3
------------
7n + 10n^1/2

You 'COMPARISON TEST' by observing that the series looks like something you have seen and whose behavior is known.  (Where are you from?  Should I be writing 'behaviour?')

So maybe this one looks like the harmonic series.  Let's try that.  We fiddle with some inequalities with this in mind:

(A) If we think this converges, prove that it is LESS than some converging series.
(B) ...............  diverges,  ---------------  MORE ......... diverging  .......

If n > 1, then we must have:

      n^1/2 < n  (square root of a number is less than the number)
    10n^1/2 < 10n
7n + 10n^1/2 < 17n

Then (note that reciprocating requires flipping the < sign.)

     3          3     3  1
------------  > --- = -- ---
7n + 10n^1/2    17n   17  n

But the series 1/n is the harmonic series and diverges.  Since the terms of your series are larger than the terms of the harmonic series, your series diverges.

- - - - - - - - - - - - - - - - - -
"If a sub n and f(n) satisfy the requirements of the Integral Test,
and if the integral from 1 to infinity of f(x)dx converges, then the summation from n=1 to infinity of a sub n = the integral from 1 to infinity of f(x)dx." I think that it's false, but i'm not sure that I fully understand what it's saying so I thought I would ask. Thank you for your help.

Your statement that the summation of a(n)  [that means a-sub-n] is equal to the integral of f(x) is not exactly true.  The integral test simply says that if the integral converges, so does the series, and vice versa, not that they give the same answer.

About the Integral Test and what it means:

Think back to those ancient days when you were first learning what an integral meant.  You drew a graph of a function on an interval, chopped your interval into pieces, then in each piece you chose an x that would give you one of these things:

(A) An upper sum, in which you have rectangles that fully enclose the area under the graph, plus extra stuff above, which --> 0 when the intervals became tiny.  Rule: the upper sum is always greater than the integral.

(B) A lower sum, in which you have rectangles that ARE fully enclosed IN the area under the graph, with extra stuff above, which --> 0 when the intervals became tiny.  The lower sum is always less than the integral.

What's the integral test?  How does it prove convergence/divergence?  Suppose you have an improper integral that converges.  That means it represents a finite area. If you have a series where the terms correspond to rectangles that make up a lower sum, then since this lower sum is INSIDE the converging integral, it must represent a finite area, therefore converge.  The reverse is true for upper sums -- if they enclose an infinite area, then they diverge.