# Interesting sums

• metacristi

#### metacristi

interesting inequality involving sums

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

&#8721k=2n {1/[k * ln(k)]} &#8805 20

Any ideas?

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1.0488269074484
Hm... I wonder how i got that.

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

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.

Originally posted by metacristi
Which is the smallest n-natural number- for this inecuation:

&#8721k=2n {1/[k * ln(k)]} &#8805 20

Any ideas?

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

very easy
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

I believe Newton1 is correct in assuming the series diverges by integral test:

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!