# analysis-bounded monotone sequence problem

• April 24th 2010, 05:40 AM
mehn
analysis-bounded monotone sequence problem
Prove that the sequence
lim n-->infinity 1/n(1 + 1/2 + ...+ 1/n) = 0

is
(i) monotone
(ii) bounded
(iii) find its limits

How can i proceed to prove that it is monotone and bounded???
Thank You!
• April 24th 2010, 08:45 AM
becko
Is 1 + 1/2 + ... + 1/n in the denominator?

If it is in the numerator, then this is divergent, since it is greater than 1/n, which is divergent.

(EDIT: nevermind this. I thought it was a sum!)
• April 24th 2010, 08:48 AM
becko
If 1 + 1/2 + ... + 1/n is in the denominator, then the denominator itself is increasing (at each step the sum gets bigger, and it gets multiplier by a greater number). This implies that the sequence is decreasing. It is bounded below since all its terms are positive. And it will converge to 0, since it is smaller than 1/n, which converges to 0.
• April 24th 2010, 08:53 AM
mehn
Thank You.
It is 1/n * (1 + ... )

• April 24th 2010, 08:54 AM
mehn
Thank You.
It is 1/n * (1 + ... )

• April 24th 2010, 08:59 AM
becko
an = (1/n) * (1 + 1/2 + ... + 1/n) < (1 + 1/4 + ... +1/n^2)

this last sequence is convergent since it is the series with term 1/k^2. This proves that your sequence is bounded.
• April 24th 2010, 09:52 AM
becko
a_n = (1/n) * (1 + 1/2 + ... + 1/n)

a_(n+1) = 1/(n+1) * (n*a_n + 1/(n+1))

(n+1) * ( a_(n+1) - a_n ) = 1/(n+1) - a_n = 1/(n+1) - (1/n) * (1 + 1/2 + ... +1/n) < 1/(n+1) - 1/n < 0

This proves that the sequence is monotone.
• April 24th 2010, 11:31 AM
becko
${a}_{n}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{k}
=\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{n} }\frac{1}{k}
<\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{{k}^{3/2}}$

The last expression tends to zero, since it is the product of a convergence sequence (the series) and an infinitesimal one. This proves that a_n tends to zero
• April 24th 2010, 01:06 PM
Plato
This is a subtopic of the problem of sequence of means.
Suppose that $(a_n\ge 0)$ is a sequence define $S_n = \sum\limits_{k = 1}^n {a_k } \,\& \,M_n = \frac{{S_n }}{n}$.
Theorem: If $(a_n)$ monotonic then $(M_n)$ is monotonic.
Theorem: If $(a_n)\to L$ then $(M_n)\to L$.