A calculus problem by jatin yadav

Calculus Level 3

If lim ⁡ n → ∞ ∑ k = 1 n ln ⁡ ( 1 + k 80 n 81 ) \displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} \ln \bigg(1+ \frac{k^{80}}{n^{81}} \bigg) equals p p . Then , find the value of 1 p \frac{1}{p}


The answer is 81.

Jatin Yadav by jatin yadav , 23, India

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1 solution

Pranav Arora
Jan 15, 2014

For the given problem, we use the approximation ln ⁡ ( 1 + x ) ≈ x \ln(1+x) \approx x . Hence, we have

lim ⁡ n → ∞ ∑ k = 1 n ln ⁡ ( 1 + k 80 n 81 ) ≈ lim ⁡ n → ∞ ∑ k = 1 n k 80 n 80 1 n \displaystyle \lim_{n \rightarrow \infty} \sum_{k=1}^n \ln\left(1+\frac{k^{80}}{n^{81}}\right) \approx \lim_{n \rightarrow \infty} \sum_{k=1}^n \frac{k^{80}}{n^{80}}\frac{1}{n}

The above is equivalent to the following definite integral:

∫ 0 1 x 80 d x = 1 81 \displaystyle \int_0^1 x^{80}\,dx=\frac{1}{81}

⇒ p = 1 81 ⇒ 1 p = 81 \displaystyle \Rightarrow p=\frac{1}{81} \Rightarrow \boxed{\frac{1}{p}=81}

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Precisely how I did it. Good Explanation.

Anish Puthuraya - 7 years, 4 months ago

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Thanks Anish! :)

Pranav Arora - 7 years, 4 months ago

Good job.

You could also multiply by n n \frac{n}{n} to get the integral without approximations.

Logan Dymond - 7 years, 4 months ago

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Thanks Logan! :)

I am not sure how can I get the integral without approximations. Can you please show a few steps?

Pranav Arora - 7 years, 4 months ago

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lim ⁡ n → ∞ ∑ k = 1 ∞ ln ⁡ ( 1 + k 80 n 81 ) = \lim_{n \to \infty} \sum_{k=1}^{\infty} \ln \left( 1 + \frac{k^{80}}{n^{81}}\right) = lim ⁡ n → ∞ ∑ k = 1 ∞ n n ln ⁡ ( 1 + k 80 n 81 ) = \lim_{n \to \infty} \sum_{k=1}^{\infty} \frac{n}{n} \ln \left( 1 + \frac{k^{80}}{n^{81}}\right) = lim ⁡ n → ∞ ∑ k = 1 ∞ ln ⁡ ( 1 + ( k n ) 80 1 n ) n 1 n = \lim_{n \to \infty} \sum_{k=1}^{\infty} \ln \left( 1+ \left(\frac{k}{n}\right)^{80} \frac{1}{n}\right)^{n} \frac{1}{n} =

Notice the definition of e e . Take the limit and you'll see that you get exactly the same integral.

Logan Dymond - 7 years, 4 months ago

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@Logan Dymond – Did it the same way. Basically Logan, your ways of writing are different but internally it's the same. The sequence you mentioned pulls down to \e\ as n → ∞ n \rightarrow \infty & the approximation works as x → 0 x \rightarrow 0 tending the same way. You can even prove this by expansion.

A Brilliant Member - 7 years, 4 months ago

Nice one

Rohan Chandra - 7 years, 4 months ago

instead of using approximation you could have used -limit as x tends to infinity ln(1+x)/x =1.

rahul saxena - 5 years, 7 months ago

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