- #1

- 252

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**interesting inequality involving sums**

Which is the smallest n-natural number- for this inecuation:

∑

_{k=2}

^{n}{1/[k * ln(k)]} ≥ 20

Any ideas?

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- Thread starter metacristi
- Start date

- #1

- 252

- 1

Which is the smallest n-natural number- for this inecuation:

∑

Any ideas?

Last edited:

- #2

- 11

- 0

1.0488269074484

Hm... I wonder how i got that.

Hm... I wonder how i got that.

- #3

- 252

- 1

1.0488269074484

Hm... I wonder how i got that.

Very interesting indeed...but how did you get that?'n' must be an integer...

In case it is not clear the sum is from k=2 to n.

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- #4

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I do not see where that number comes into play. It is certainly not the answer.

- #5

- 11

- 0

Originally posted by enslam

1.0488269074484

Hm... I wonder how i got that.

I crawl back into my hole and will read the question properly next time.

- #6

- 204

- 1

Originally posted by metacristi

Which is the smallest n-natural number- for this inecuation:

∑_{k=2}^{n}{1/[k * ln(k)]} ≥ 20

Any ideas?

Nonexistent.

I think if we let n go to infinity we'll still never get above 5 for the sum.

- #7

- 152

- 0

u just use the integral test

because the sum is continuous and decreasing

if n approach to infinite

the sum is diverges and must greater than 20

- #8

suffian

Let f(x) = 1/( x ln(x) ), then since f(x) is positive, continous, and decreasing for n>=2 we apply Integral test.

Integral[ 2->inf, 1/(x lnx) ] diverges, therefore the series diverges as well.

Since the partial sums in this series are non-decreasing (all the terms in this series are positive) , then there will exist a least natural number N for which the sum will exceed 20 (and then stay above it).

I'm not sure what the value of N is, but by applying a variant on the integral test, I think i've found a real value B for which N must be greater.

Suppose you approximate the following integral with right-hand riemann sums with delta_x = 1:

Integral[ 2->B, 1/(x lnx) ]

Then the riemann sum approximation will be the sum from k=3 to B of 1/ ( n ln(n) ) (assuming B is integer) and this approximation will be an underestimate.

From which it follows:

Sum[k=2->B, 1/ (n ln(n) )] <

Integral[2->B, 1/ (x lnx )] + 1/(2ln2)

The value of N must be greater than the value of B that first causes the integral + 1/(2ln2) to go over 20 because the sum will be less than the integral + 1/(2ln2).

Integral[ 2->B, 1/(x lnx) ] + 1/(2ln2) >= 20

ln(ln(B)) - ln(ln(2)) +1/(2ln2) >= 20

B >= e^e^(20- 1/(2ln2) +ln(ln(2)))

B >= 2.726413 * 10^70994084

This value is beyond astronomical!

And if I've done my calculations correctly, N must be greater than this!

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