Limit of a sequence (Calculus 1)

Why is the following statement True ?

limn→+∞ an = L then limn→+∞ 1/n Σk=1n ak = L

If you're using epsilon-delta definitions, here's a proof:

For e>0 there's an N such that for k>N, |ak - L|<e/2.

Take M = |(a1-L) + (a2-L) + .... + (aN-L)|, then for k>N

|a1+...+ak - k*L| <= M + (k-N)e/2

dividing by k

|(a1+....+ak)/k - L| <= M/k + ((k-N)/k)(e/2)

and in the limit k->inf that's e/2, so there's some N' where k>N' implies that's <e.

That means for the given e>0, that N' satifies the limit definition for

lim (a1+...+ak)/k = L

Integer of (1+x²)/abs(1-x²)

If 1 - x² => 0 :

int ( 1 + x² ) / | 1 - x² | dx

= int ( 1 + x² ) / ( 1 - x² ) dx

= int ( x² - 1 + 2 ) / ( 1 - x² ) dx

= int -1 + 2 / ( 1 - x² ) dx

= int -1 + 2 / [(1 - x)(1 + x)] dx

2 / [(1 - x)(1 + x)] = a / (1 - x) + b / (1 + x)

=> 2 = a(1 + x) + b(1 - x) = a + ax + b - bx = (a - b)x + a + b

=> a = b and a + b = 2

=> a = 1 and b = 1

So :

int ( 1 + x² ) / | 1 - x² | dx

= int -1 + 1 / (1 - x) + 1 / (1 + x) dx

= int -1 - (-1) / (1 - x) + 1 / (1 + x) dx

= [-x - ln(|1 - x|) + ln(|1 + x|)]

Can you do 1 - x² < 0 ?

For the subject Becquerel(Bq) to disintegerate from 11.5Mbq to 4.6 MBq takes 8 hours. How much percent does it disintegerate per hours?And how big is it after a 24 hours if it starts from 11.5 MBq?

Final = Initial*e^rt

4.6 = 11.5e^8r

4.6/11.5 = 2/5 = e^8r

ln(2/5) = ln(2) - ln(5) = 8r

(ln(2) - ln(5))/8 = r ~ -11%/hour.

To find how much is left after 24 hours:

Final = 11.5e^{[24(ln(2)-ln(5))/8]} = 11.5e^{[3(ln(2)-ln(5))]} = 11.5(2/5)³