Expert: Paul Klarreich - 3/1/2010

Question
Sorry to have made these questions private, but if this site works and I really hope it does as I really need help, I have follow up questions which will not be private.

(1) Determine, with proof, whether the following series converge or not:
(a)
∞
∑  (2n)!(3n)! / n!(4n)!
n=1

(b)
∞
∑ 1/n^[1+(1/n)]
n=1
I started by identifying that n^[1+(1/n)] = (n)*[n^(1/n)], and I think I remember proving that n^(1/n) lim → e, but not sure what to do.
(2)I started by proving that

lim    [ 1/n + 1/(n+1) +…+ 1/(2n) ] = ln2
n→1

Now I must start with a CONVERGENT series
∞
∑    a_n (a sub n or a with a base n) instead of the divergent   
n=1   series ∑ (1/n)

and we let:   b_n = a_n + a_(n+1)+… +a_(2n).

Prove that
lim   b_n = 0.
n→∞
I know I need to show: b_n = 1/n^2 + 1/(n+1)^2  + 1/(n+2)^2 + ... + 1/(2n)^2 but not sure how.

(3)This question I wasn't able to do much with but I have to prove that if a_n → 0, then the sequence
                     
                  b_n = (a_1 + a_2 +…+ a_n ) / n

also converges to 0. Remember that a_n→ 0 tells you nothing about, say, a_1 or a_2, so when you estimate b_n you will have to separate the early a_n's from the late ones.

Answer
Questioner: Sarah
Country: United States
Category: Advanced Math
Private: Yes
Subject: 3 analysis question to finish my study guide

Question: Sorry to have made these questions private,

>> No problem -- I fixed it for you.

but if this site works and I really hope it does as I really need help, I have follow up questions which will not be private.

(1) Determine, with proof, whether the following series converge or not:

(a)
∞
∑  (2n)!(3n)! / n!(4n)!
n=1

Looks like a ratio test example, I think:

a[n+1]/a[n]   << notation for subscripts.

= (2n+2)!(3n+3)!   n!(4n)!
 -------------- ----------
  (n+1)!(4n+4)! (2n)!(3n)!
 2n+2 2n+1  3n+3  3n+2  3n+1   << () omitted
= ----------------------------
 n+1  4n+4  4n+3  4n+2  4n+1

now lim(n->inf) of that would be:
 2 2 3 3 3     36
------------ = ----- < 1, which is good.  It converges.
 1 4 4 4 4     256

.................................

(b)
∞
∑ 1/n^[1+(1/n)]
n=1
I started by identifying that n^[1+(1/n)] = (n)*[n^(1/n)], and I think I remember proving that n^(1/n) lim → e, but not sure what to do.

>> I don't remember that.  I think  lim n^(1/n) = 1.

ln [n^(1/n)] = ln n / n, which --> 0. so n^(1/n) --> e^0 = 1
................
You are correct in that:

     1
------------ =
n^(1 + 1/n)

1     1
- ---------- -->  1/n.
n   n^(1/n)

So as n-> inf, your sum looks like  1/n, the harmonic series, which diverges.
----------------------------

(2)I started by proving that

lim    [ 1/n + 1/(n+1) +…+ 1/(2n) ] = ln 2
n→1

Now I must start with a CONVERGENT series
∞
∑    a_n (a sub n or a with a base n) instead of the divergent   
n=1   series ∑ (1/n)

and we let:   b_n = a_n + a_(n+1)+… +a_(2n).

Prove that
lim   b_n = 0.
n→∞

I know I need to show: b_n = 1/n^2 + 1/(n+1)^2  + 1/(n+2)^2 + ... + 1/(2n)^2 but not sure how.

>> I don't think that is relevant.  You should be able to prove that in general, if  SUM a[n] converges, then your b[n] --> 0.

And I think it could go like this:

Suppose SUM a[n] converges to A.  Call the partial sums A[n],
where  A[n] = SUM(1 to n) a[k]

Then given e,   (I cannot make epsilons;  e will have to do.)
we can find N such that

A - A[k] < e

But  b[N] (as defined above) is <= A - A[k], so it is approaching zero.

That should do it.

...........................

(3)This question I wasn't able to do much with but I have to prove that if a_n → 0, then the sequence
                    
                 b_n = (a_1 + a_2 +…+ a_n ) / n

also converges to 0. Remember that a_n→ 0 tells you nothing about, say, a_1 or a_2, so when you estimate b_n you will have to separate the early a_n's from the late ones.

((Note that we don't know if  SUM a[n] converges.))

yes, you split into two parts.

You want

(a_1 + a_2 +…+ a_n ) / n  <  e

Now there exists N such that a[k] < e1 for  n > N.

Split  a[1] + .. + a[n] into:


a[1] + .. + a[k-1] PLUS a[k] + .. + a[n]

Look at the first part:  a[1] + .. + a[k-1].  This is a finite sum.  Suppose its sum is B.

Look at the second part:  a[k] + .. + a[n].  The number of terms in this varies with n, but we know that:

1) there are fewer than 'n' of these terms.
2) each is <= e1.

So we make choices: (sorry if this sounds like a drug commercial)

Choose 'n' so that B/n < e/2

Choose e1 so that  n*e1 < e/2.  I.e. make e1 = e/2n

Now the two sums are less than  e.

Does this make any sense?