SOLUTION: How do you solve 1/2+2/3+3/4+4/5......+2013/2014?

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Question 1010159: How do you solve 1/2+2/3+3/4+4/5......+2013/2014?
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
1/2+2/3+3/4+4/5......+2013/2014 =
That has 2013 terms, and is equal to

(1-1/2)+(1-1/3)+(1-1/4)+(1-1/5)+...+(1-1/2014) =

(1+1+...+1)-1/2-1/3-1/4-1/5-...-1/2014) =

There are 2013 1's added so this is:

2013 - (1/2+1/3+1/4+1/5+...+1/2014)

That series in parentheses starts with 1/2.
The harmonic series starts with 1/1, so let's
add and subtract 1:

which equals:

2013 + 1 - 1 - (1/2+1/3+1/4+1/5+...+1/2014) =

2014 - (1+1/2+1/3+1/4+1/5+...+1/2014)

In the parentheses in the harmonic sequence to 2014 terms.

There is no formula for that, but a good approximation for it is:

Hn = ln(n) + g + 1/(2n) - 1/(12n^2)

where g = the Euler-Mascheroni constant, which
is approximately 0.5772156649.  You can Google information on that.

Substituting n = 2014

H2014 is approximately 8.185342021, making the desired sum
approximately

2014 -  8.185342021 = 2005.814658

When I ran that sum on the computer, I got 2005.81466

Edwin

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