SEQUENCE AND SERIES

[Eco16]

sigma n=1 to n= infinty n!/n^n ?
Sigma n=1 to n= infinity ln n/n^2?
Please help!

[Eco16]

TEST WHETHER THEY CONVERGE OR DIVERGE?

[Dr. Strange]

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.

[samuraijack]

for question 1, Ratio test will help,

 if limit  |Tn+1/Tn| < 1 series converges
    n-infinity

if we do this we get limit (n/n+1)^n which is 1/e < 1 series converges

[Dr. Strange]

Thanks

[Tania123]

For 1, We can also use the nth term test.
Proving lim(n tends to infinity) a(n) =0 we can prove that the series is convergent.

[Dr. Strange]

The nth-term test for divergence is a  test for the divergence of an infinite series.
Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. In particular, the converse to the test is not true; instead all one can say is:

[Eco16]

thank you!!
Sigma n=1 to infinity 3n+4/2n^2+3n+5? using comparison test check if it converges or diverges.

Can somebody help me with this?

[Eco16]

also n=1 to infinity 3^n/2^n+3^n? comparison test!

[Akash]

[Akash]

[Eco16]

isn't ratio test defined as a n+1/a n?

[Mauli]

I think the ratio test will fail in this question. We will have to apply Rabe's test.

PS: this is just conjecture.

[Akash]

Yeah sry. I meant the limit comparison test  Its the one where you look at the limit for some related series which you know diverges/converges. If its a finite constant, then the series in question will do the same too.



And Raabe's works too

[Old monk]

what is the sum of x+(1/2*X^2)+(1/3*x^3)+........?

[Akash]

-ln(1-x) has the same series expansion for x belongs to (-1,1) and infinite series

[Old monk]

Thanks :)