Is there any mathmatical solution of hermonic serise(1+1/2+1/3+...+1/n)where n is known?

Answer 1

Of course.
if n is known, say n=4 then,
1+1/2+1/3+1/4 =2.08333...

If you are asking in terms of a Limit to the series then

Lim [1+1/2+1/3+...+1/n] ~ 2 1/2

Answer 2

The harmonic series diverges. So if n = ∞. The sum = ∞.

This can be seen this way.

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) ...
< 1 + 1/2 + (1/4 + 1/4) +(1/8 + 1/8 +1/8 +1/8) ...
= 1 + 1/2 + 1/2 + 1/2 ...= ∞

Because you are adding an infinite number of 1/2's together.

Answer 3

Trace function y(x)=1/x between x1=j and x2=j+1; the area of trapezium between y(x) and x-axis s(j) = integral{for x=j until j+1} of dx/x = ln(x){for x=j until j+1} = ln(j+1) –ln(j); meanwhile a trapezium s(j) < rectangle 1/j;
Thus H(n) = Sum{for j=1 until n} of 1/j > Sum{for j=1 until n} of {ln(j+1) -ln(j)} = (ln2-ln1) +(ln3-ln2) +(ln4-ln3) +(ln5-ln4) +++ (ln(n+1)-ln(n)) =ln(n+1) = or = integral{for x=1 until n+1} of dx/x;
If n - -> infinity then ln(n+1) - - > infinity; so H(n) - - > infinity for sure!
This was the proof that harmonic series does not converge.
Now a formula: suppose you know exactly E(5) = 1+1/2 +1/3 +1/4 +1/5 =2.283333; then H(10) = E(5) +ln(10.5/5.5) = 2.92996 will be a good approximation for exact E(10) = 2.928968;
Approximate H(20) = E(5) +ln(20.5/5.5) = 3.59901, while exact H(20) = 3.59774;
Or a better approximation H(20) = E(10) +ln(20.5/10.5) = 3.598018;
I’d recommend H(n) =E(10) +ln((n+0.5)/10.50292) for better coincidence.

Answer 4

There is no nifty formula to calculate it exactly. To calculate it exactly will take some brute work.

Answer 5

well the denominator
would be n!
and the numerator
wud be something like:
n!+n!/2+n!/3+......+1
i think

Answer 6

NO POSSIBLE ANSWER.SORRY