Limit of a sequence

Calculus Level 5

Let us define $k_0=0$ and $k_{n+1}=k_n+\sqrt{1+k_n^2}$ .

Find the value of $\lim_{n \rightarrow \infty} \left(\dfrac{k_n}{2^n}\right)$

Also try Limit of a sequence (2) and (3)

$\textbf{Assumptions and Source}$

$n$ is any non-negative integer.
This is a standard problem in most calculus textbooks for $\text{JEE (Advanced)}$ .

The answer is 0.6366.

1 solution

Ronak Agarwal
Sep 8, 2014

Assume ${ k }_{ 0 }=cot({ \theta }_{ 0 })$ . So we have

${ k }_{ 1 }=cosec({ \theta }_{ 0 })+cot({ \theta }_{ 0 })=\frac { 1+cos({ \theta }_{ 0 }) }{ sin({ \theta }_{ 0 }) } =\frac { 2{ cos }^{ 2 }(\frac { { \theta }_{ 0 } }{ 2 } ) }{ 2sin(\frac { { \theta }_{ 0 } }{ 2 } )cos(\frac { { \theta }_{ 0 } }{ 2 } ) } =cot(\frac { { \theta }_{ 0 } }{ 2 } )$

Similarly we have :

${ k }_{ 2 }=cot\frac { { \theta }_{ 0 } }{ { 2 }^{ 2 } }$

Continuing like this we have :

${ k }_{ n }=cot(\frac { { \theta }_{ 0 } }{ { 2 }^{ n } } )$

Now back to the limits we have :

$\lim _{ n\rightarrow \infty }{ \frac { { k }_{ n } }{ { 2 }^{ n } } } =\lim _{ n\rightarrow \infty }{ \frac { cot(\frac { { \theta }_{ 0 } }{ { 2 }^{ n } } ) }{ { 2 }^{ n } } } =\lim _{ n\rightarrow \infty }{ \frac { 1 }{ { 2 }^{ n }tan(\frac { { \theta }_{ 0 } }{ { 2 }^{ n } } ) } } =\frac { 1 }{ { \theta }_{ 0 } }$

In our given question $cot({\theta}_{0})=0$

$\Rightarrow{\theta}_{0}=\frac{\pi}{2}$

Our limit becomes $L=\frac{2}{\pi}$

Done the same way

Ayush Garg - 6 years, 9 months ago

@Ronak Agarwal what was the logic in assuming ksub0 as cot theta can u explain

Tejas Suresh - 6 years, 8 months ago

The hint lies in that $\sqrt {1+{{k}_{n}}^{2}}$ .

Arif Ahmed - 6 years, 8 months ago

Just see the form of the recurrence putting ${k}_{0}=cot(\theta)$ helps a lot as you can see in the question.

Ronak Agarwal - 6 years, 8 months ago

We can also substitute k1 as tanα to get the same answer 2/π.

Mayank Jha - 4 years, 2 months ago

Great proof, yet I'm still a bit confused on that last limit evaluated to 1/theta. Can you please tell me how?

Sadat Shaik - 6 years, 9 months ago

He calculated $\frac 1 {\theta}_{0}\;$ using "l'Hôpital's rule": $\lim_{n \to \infty} \frac {2^{-n}}{\tan{(2^{-n} \theta}_{0})} =$ $=\lim_{n \to \infty} \frac {\frac d {dn}2^{-n}}{\frac d {dn}\tan{(2^{-n} \theta}_{0})} =$ $=\lim_{n \to \infty} \frac {-n2^{-n-1}\ln2}{\frac {-n2^{-n-1}\theta_{0}\ln2}{{cos}^2(2^{-n}{\theta}_{0})}} =$ $= \frac {{cos}^2(0)} {{\theta}_{0}} = \frac 1 {\theta}_{0}$

Antonio Fanari - 6 years, 8 months ago

Ah! I put thera(0) as pi .

Aakash Khandelwal - 5 years, 7 months ago

Very Good Sum and Solution!!!!Did the same way :)(+1) obv...!!!!

rajdeep brahma - 3 years ago

but then there are infinite no of solutions as 3pi/2 is also a valid solution thus shouldn't the limit be non existant????

Ashish Mohanty - 2 years, 3 months ago

Limit of a sequence

The answer is 0.6366.

1 solution

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