Routine limes

Calculus Level 4

Find

X = lim ⁡ n → ∞ 1 k + 2 k + 3 k + ⋯ + n k n k + 1 X=\lim_{n \to \infty} \frac{1^k+2^k+3^k+\dots+n^k}{n^{k+1}}

Put the answer for k=78 as the 100*X.


The answer is 1.265.

Вук Радовић by Вук Радовић , 25, Serbia

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1 solution

Using Stolz–Cesàro theorem we have:

X = lim ⁡ n → ∞ ( 1 k + 2 k + 3 k + ⋯ + ( n + 1 ) k ) − ( 1 k + 2 k + 3 k + ⋯ + n k ) ( n + 1 ) k + 1 − n k + 1 X=\lim_{n \to \infty} \frac{(1^k+2^k+3^k+\dots+(n+1)^k)-(1^k+2^k+3^k+\dots+n^k)}{(n+1)^{k+1}-n^{k+1}}

X = lim ⁡ n → ∞ ( n + 1 ) k ) ( n + 1 ) k + 1 − n k + 1 X=\lim_{n \to \infty} \frac{(n+1)^k)}{(n+1)^{k+1}-n^{k+1}}

Bernoulli's inequality we can find:

lim ⁡ n → ∞ n k + 1 ( 1 + 1 n ) k + 1 − n k + 1 = lim ⁡ n → ∞ n k + 1 k + 1 n \lim_{n \to \infty} n^{k+1}(1+\frac{1}{n})^{k+1}-n^{k+1}=\lim_{n \to \infty} n^{k+1}\frac{k+1}{n}

And now X = 1 k + 1 X=\frac{1}{k+1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Alternatively we could convert into a Riemann sum Forming the integral of x^78 dx With limits from 0 to 1 This quickly yields I=1/79 And thus 100/79=1.265 approx

Suhas Sheikh - 3 years ago

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