14 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.only thing i can figure out is writing 1/r+3 as integral x^2r from 0 to 1
@SirLancelotDuLac
Ooh this looks fun...
Ye, I have an idea:
Write the numerator as $\frac{r+3-r}{3}(\text{stuff})$ then
$\sum{r=0}^{n} \frac{\binom{n}{r}}{3.n^{r}}-\sum{r=0}^{n}\frac{\binom{n}{r}r}{3.n^{r}(r+3)}$
After this, let the limit be L
Then note that as n tends to infinity
SirLancelotDuLac
the second term in the expression above tends to L/3 and the first term is (1+1/n)^n/3
So L=e/3-L/3 which means 4L/3=e/3
which means L=e/4?
I'm not sure, I'll look at this again if it's incorrect.
Though is the answer e/4? @hardcoreisdead
e-2
Did you ever get an answer to this?
If so, can you please share in this thread before closing it?
i kinda did
i am not sure of step 3 tho
Gahh I missed that.
Step 3 is just limit(a tending to something) integral(expression of x and a).dx is integral of limit of (expression of x and a).dx
oh right
damn W ques