find-for-arbitrary-0-the-limit-lim-n-1-2-3-n-n-n-1-

Question Number 169166 by infinityaction last updated on 25/Apr/22

find for arbitrary α ⩾ 0 the limit

\lim_{n \to + \infty} (\frac{1^{α} + 2^{α} + 3^{α} + \dots \dots n^{α}}{n^{α}} - \frac{n}{1 + α})

Commented by safojontoshtemirov last updated on 25/Apr/22

\frac{1}{2}

Commented by infinityaction last updated on 25/Apr/22

s i r s o l u t i o n

Commented by infinityaction last updated on 25/Apr/22

s i r d o n t p u t a n y v a l u e o f α

Commented by infinityaction last updated on 25/Apr/22

i w a n t s o l u t i o n i n g e n e r a l f o r m

Answered by aleks041103 last updated on 25/Apr/22

\int_{1}^{n + 1} x^{α} d x = \frac{{(n + 1)}^{α + 1} - 1}{α + 1} \approx (\underset{k = 1}{\sum^{n}} k^{α}) + \underset{k = 1}{\sum^{n}} (\frac{1}{2} ({(k + 1)}^{α} - k^{α})) =

= (\underset{k = 1}{\sum^{n}} k^{α}) + \frac{1}{2} ({(n + 1)}^{α} - 1)

\Rightarrow \underset{k = 1}{\sum^{n}} k^{α} \approx (\frac{n + 1}{α + 1} - \frac{1}{2}) {(n + 1)}^{α} + \frac{1}{2} - \frac{1}{α + 1}

\Rightarrow \underset{n \to \infty}{l i m} ((\frac{n + 1}{α + 1} - \frac{1}{2}) {(1 + \frac{1}{n})}^{α} - \frac{n}{1 + α} + (\frac{1}{2} - \frac{1}{α + 1}) \frac{1}{n^{α}}) =

= \underset{n \to \infty}{l i m} ({(1 + \frac{1}{n})}^{α} (\frac{1}{α + 1} - \frac{1}{2}) + \frac{n}{α + 1} ({(1 + \frac{1}{n})}^{α} - 1)) =

= \frac{1}{α + 1} - \frac{1}{2} + \frac{α}{α + 1} = \frac{1}{2} f o r α > 0

f o r α = 0

\Rightarrow l i m = \frac{1 + 1 + \dots + 1}{1} - n = 0

\Rightarrow \underset{n \to \infty}{l i m} (\frac{1^{α} + \dots + n^{α}}{n^{α}} - \frac{n}{α + 1}) = {\begin{cases} 0, α = 0 \\ 1 / 2, α > 0 \end{cases}

Commented by infinityaction last updated on 25/Apr/22

t h a n k y o u s i r