Re: SEQUENCE AND SERIES

Posted by Dr. Strange on Apr 16, 2016; 12:27pm
URL: http://discussion-forum.276.s1.nabble.com/SEQUENCE-AND-SERIES-tp7599873p7599882.html

2.Sigma n=1 to n= infinity ln n/n^2

Sum= ln 1/1 + (ln2)/ 4 .... = (ln 2)/4+ (ln 3)/9 and so on

let f(x)= ln x /x^2

 Since f is a monotone decreasing function and  integration of ln x /x^2 from x=1 to infinity converges to 1 so the sum converges to 1 as well

(by Integral test for convergence)

You can check this link :  https://en.wikipedia.org/wiki/Integral_test_for_convergence

1. sigma n=1 to n= infinty n!/n^n
 
Sum= 1/1 + 2/2^2  + 3*2/ 3^3 + 4*3*2/ 4^4 and so on

If we simplify a lil,
Sum= 1+ 1/2 + 2/9 + 2*3/ 4^3 .....

We exclude 1 for the moment and take series 1/2 + 2/9 + 6/64...
 
New sum = 1/2 + 2/9 + 2*3/ 4^3....

We can see T(n+1)= T(n) *  (n+1/n+2)^ (n+1) from n=1 onwards

So T(n) gets multiplied by  (n+1/n+2)^ (n+1) to get T(n+1)

Now the multiplier is a decreasing function as (x+1/x+2) ^ (x+1) is decreasing with maximum value at x=1
So T(n) *  (n+1/n+2)^ (n+1)< T(n) * (4/9)
So T(n+1) < T(n) * (4/9)

New sum is less than a g.p series with first term 1/2 and common ratio 4/9
so new sum < 9/10

so sum < 1+ (9/10)

sum < 19/10

If you find a quicker method for ques 1 please share it.