Re: Proving lim n * a_n = 0
- From: rob@xxxxxxxxxxxxxx (Rob Johnson)
- Date: Thu, 01 Feb 2007 21:50:10 GMT
In article <1170331041.181528.93600@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
nicegirl_130@xxxxxxxxx wrote:
I started studying sequences and series abou 2 months ago and I'm not
very secure about some points yet, so I'd like someone to check my
proof. I still have some difficulty to deal with the concepts of lim
inf and lim sup of a sequence. I used to deal more with integers and
prime numbers, so I have some difficulty with limit processes.
Show that, if (a_n) is a motonic sequence of real numbers and Sum(n=1,
oo) converges, then lim n a_n = 0.
First, let's suppose (a_n) is decreasing. Then, lim a_n =0 and a_n >=0
for every n, since infimum (a_n) = 0. For each natural k, let S_k =
Sum(n=k, oo) a_n. Since Sum(n=1, oo) converges, each S_k is finite
and lim S_k =0.
Keeping k fixed and taking into account (a_n) is decreasing, for every
n >=k we have 0 <= a_n ....+ a_n = (n -k +1) a_n <= a_k ....+ a_n <=
S_k. Therefore, 0 <= n a_n <= (k-1) a_n + S_k. Since this holds for
every n >=k (that is, with possibly exception of a finite number of
values of n, it holds for all n) it follows that 0 <= lim sup n a_n <=
lim sup ((k-1) a_n + S_k.). Since a_n -> 0, ,(k-1) a_n + S_k.) ->
S_k. For this sequence, lim inf, lim and lim sup are therefore the
same number. This leads to 0 <= lim sup n a_n <= S_k. Since these
inequalities hold for every k in {1,2,3....}, we must have 0 <= lim
sup n a_n <= lim S_k = 0 => lim sup n a_n = 0.
Since n a_n >= 0 for every n, we also have 0 <= lim inf n a_n < =
limsup n a_n = 0 . Therefore lim inf and lim sup are both 0 , and we
simply have lim n a_n =0.
If (a_n) is increasing, we can use a similar reasoning. But since we
already have the previous conclusion, it's enough to apply it to the
decreasing sequence (- a_n) to conclude that lim (n * (-a_n)) = lim (-
n a_n) = -lim (n a_n) = 0, so that lim n a_n =0.
Based on this conclusion, I found an easy way to prove the harmonic
series diverges. If a_n =1/n, then, for every n, n a_n =1 and lim n
a_n =1, so that the condition lim n a_n =0, necessary for convergence,
is not satisfied. Therefore, Sum (1/n) diverges. I hadn't seen this
proof before. I agree it's not very informative, other proofs show
much better why this series goes to infinity.
Another way is to note that
oo
---
a = > a - a [1]
n --- k k+1
k=n
Since the following deal with sums of positive numbers, we can
change the order of summation:
oo
---
T = > a
N --- n
n=N
oo oo
--- ---
= > > a - a using [1]
--- --- k k+1
n=N k=n
oo k
--- ---
= > > a - a change order of summation
--- --- k k+1
k=N n=N
oo
---
= > (k-N+1)(a - a )
--- k k+1
k=N
oo
---
= > (k+1)(a - a ) - N a [2]
--- k k+1 N
k=N
By [2] we have
oo oo
--- ---
> a = > (k+1)(a - a ) [3]
--- n --- k k+1
n=0 k=0
Therefore, we know that the sum on the right of [3] converges.
Since the sum of a_n converges, T_N -> 0 as N -> oo. Also,
oo
---
> (k+1)(a - a )
--- k k+1
k=N
tends to 0 as N -> oo since the sum on the right of [3] converges.
Putting these two limits together with [3], we get that N a_N -> 0.
Rob Johnson <rob@xxxxxxxxxxxxxx>
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