FTC - spaces between curves

nikniklau · **Joined:** Fri, 11 Nov 2011 14:43:25 UTC **Posts:** 18

Prove that $ln(n+1) \leq \sum_{n=1}^k {1 \over k} \leq ln(n+1)+{{n} \over {n+1}}$

provided that $\int_n^{n+1} {1 \over x}dx = ln({n+1 \over n})$ (this is the space between curve 1/x, x axis, x=n and x=n+1 as I calc it)

This is about the FTC. Any hints?

outermeasure · **Posted:** Fri, 13 Apr 2012 14:43:12 UTC

nikniklau wrote:

Prove that $ln(n+1) \leq \sum_{k=1}^n {1 \over k} \leq ln(n+1)+{{n} \over {n+1}}$

provided that $\int_n^{n+1} {1 \over x}dx = ln({n+1 \over n})$ (this is the space between curve 1/x, x axis, x=n and x=n+1 as I calc it)

This is about the FTC. Any hints?

No. It is about estimating integral from above and below (i.e. upper and lower Riemann sum, for suitable partition P).

nikniklau · **Joined:** Fri, 11 Nov 2011 14:43:25 UTC **Posts:** 18

outermeasure wrote:

nikniklau wrote:

Prove that $ln(n+1) \leq \sum_{n=1}^k {1 \over k} \leq ln(n+1)+{{n} \over {n+1}}$

provided that $\int_n^{n+1} {1 \over x}dx = ln({n+1 \over n})$ (this is the space between curve 1/x, x axis, x=n and x=n+1 as I calc it)

This is about the FTC. Any hints?

No. It is about estimating integral from above and below (i.e. upper and lower Riemann sum, for suitable partition P).

so the lower sum is ln(n+1)

and the upper sum is $ln(n+1)+{{n} \over {n+1}}$ and I have to figure out what the partition is.

That is to find F(b)-F(a)=ln(n+1)

for

and $ln(n+1)+{{n} \over {n+1}}$ for max f(x)

when x in [a,b].
I can find that trying but what is a suitable method for it? It is clear what max and min is for this function. Since x>0 and a<b then f(a)=max and f(b)=min.

outermeasure · **Posted:** Sat, 14 Apr 2012 01:58:53 UTC

Read the

Quote:

(...) provided that $\int_n^{n+1} \frac1x\,\mathrm{d}x = \log_e(\frac{n+1}{n})$

given in the question. It is there for a reason!

nikniklau · **Joined:** Fri, 11 Nov 2011 14:43:25 UTC **Posts:** 18

$\sum_{n=1}^k {1 \over k} = k{1 \over k}=1$ . That is the space between line $y={1 \over k}$ , x axis, lines x=1 and x=k (dk=1). How is this connected to the other parts of the inequality? ln(n+1)>1 when n>e...

outermeasure · **Posted:** Sat, 14 Apr 2012 15:19:51 UTC

nikniklau wrote:

$\sum_{n=1}^k {1 \over k} = k{1 \over k}=1$ . That is the space between line $y={1 \over k}$ , x axis, lines x=1 and x=k (dk=1). How is this connected to the other parts of the inequality? ln(n+1)>1 when n>e...

I think you want [tex]\sum_{k=1}^n\frac{1}{k{/tex] instead.

nikniklau · **Joined:** Fri, 11 Nov 2011 14:43:25 UTC **Posts:** 18

Oops...Since dk=1 then ${b-a \over k} = 1$ this can be $\int_a^{n+a} {1 \over k} dk=ln(n+a)-ln(a)=ln({n+a \over a})$
This is ln(n+1) for a = 1 and the inequality stands (as an equality on the left). And then what? Proof by induction?
$$ \displaystyle \lim_{a \to \infty}ln({n+a \over a})=ln(1)<ln(n+1), \forall n \in \mathbb{N}$

nikniklau · **Joined:** Fri, 11 Nov 2011 14:43:25 UTC **Posts:** 18

Some videos over utube and my knowledge over Riemann sums is kind of better now.
The above sum is the left Riemann Sum. Its first term is 1 and that means f(a)dx = 1 and since dx=1 then f(a)=1 and a=1.
$f'(x)=-{1 \over x^2}<0, \forall x \in D_f$ , that means that the function is monotonically decreasing and the left sum is greater that the right sum. I think the inequality is in a good way now, since we can calculate the difference between the left and the right sum.

Thanx for your help...

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FTC - spaces between curves

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