# Doubt from limits

#1
#2

getting 0 please see hari shankar sir.....

#3

Is the answer equal to 1

#4

no idea..post solution

#5

Here since we are not given the sequence the best and easiest sequence one can think of is that of ai=i
But substituting that we get n/ln(n) as n tends to infinity. But this Limit is not defined.
But again the starting is just an assumption. And we can't definitely rely on them.
How did you got Zero ? @Sneha_2021

#6

by cesaro theoram @Mayank_2019_1 sir..

1 Like
#7

@Sneha_2021
Do you know the answer for the problem?
If yes, plz post it, otherwise we cannot confirm the validity of the solution no.

Also, @Sneha_2021 in the question it is given that : \lim_{n\to\infty}a_n=n.
Do you think it makes sense? a_n is supposed to be a function of n only, and if we apply the limit on the former when the latter tends to infinity, how can we get answer in the form of n?

#8

zero is correct answer @Churchil_2018 sir..

1 Like
#9

If zero is the correct answer, you got it no?

#10

ya sir can you share your aporoch to @Churchil_2018 sir

#11

I also proved it using the stolz-cesazro theorem, do you need the solution?
@Sneha_2021
I guess thats only the plausible way of finding the answer.

1 Like
#12

Okay @Sneha_2021 i will tell you the approach and solution.

So as per the Stolz- Cesaro Theorem, let we have any sequence {b_n}, such that it obeys two properties :

\to b_{n+1}-b_n>0 \space\space\space\space\space\space(1)

and \to lim_{n\to\infty}\sum_{k=1}^n b_k=\infty \space\space\space\space\space\space(2)

and let we have any sequence {a_n} such that : lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=L(say), and lim_{n\to\infty}\frac{a_n}{b_n} exists, then the latter limit is also equal to L.

So here, let we have {b_n}=n and {a_n}=a_n as per given.

Now, we see that {b_n} satisfies the aforementioned properties 1 and 2. And hence it is eligible for the computation of the Stolz-Cesaro Limit.

Now, given that : \lim_{n\to\infty}a_n=n (even though I find it really weird, still since its given, so i am answering it).

We have : L=lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}

\implies L=\frac{\lim_{n\to\infty}(a_{n+1})-\lim_{n\to\infty}(a_n)}{\lim_{n\to\infty}(b_{n+1}-b_n)}

\implies L=\frac{(n+1)-(n)}{1} \implies L=1 \space\space\space\space\space(3)

Thus, we have obtained the value of L.

Now, the last part needs to check the existence of lim_{n\to\infty}\frac{a_n}{b_n}.

Let l= lim_{n\to\infty}\frac{a_n}{b_n}

\implies l=\frac{lim_{n\to\infty}a_n}{lim_{n\to\infty}b_n} (as individually both limits exist)

\implies l exists (utilizing the given fact that \lim_{n\to\infty}a_n=n).

Eventually we can see that l comes out to be equal to L from evaluation as well.

Thus, for any k\space\epsilon N, we have : \lim_{k\to\infty}\frac{a_k}{k}=\lim_{k\to\infty}\frac{a_k}{b_k}=1

And hence, \lim_{k\to\infty}\frac{\sum_{k=1}^n\frac{a_k}{k}}{ln(n)}=0.

Hence, the answer should be 0.

(as the numerator will remain finite, seeing to the fact that the limit is not on k, but on n).

Do let me know if you have any doubts.