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I wanted to find the limit: $\lim_{n \to +\infty}(\frac{1^p + 2^p + ... + n^p}{n^p} - \frac{n}{p + 1})$, where $p \in \mathbb{N}$.

I saw that I could use Stolz–Cesàro theorem, where $x_n = (p + 1)(1^p + 2^p + ... + n^p) - n^{p + 1}$, and $y_n = n^p(p + 1)$ tends to +$\infty$.

So, here we are:

$\lim _{ x\rightarrow \infty }{ \frac { { x }_{ n }-{ x }_{ n - 1 } }{ { y }_{ n }-{ y }_{ n - 1 } } } = \lim _{ x\rightarrow \infty }{ \frac { ( p+1 ) { n }^{ p }-{n}^{ p+1 }+{ (n-1) }^{ p+1 } }{ ( p+1 ) ( { n }^{ p }-{ (n-1) }^{ p } ) } } = \lim _{ x\rightarrow \infty }{ \frac { ( p+1 )-n+(n-1)(\frac{n - 1}{n})^p }{ ( p+1 ) ( 1-{ (\frac{n-1}{n}) }^{ p } ) } } = \lim _{ x\rightarrow \infty }{ \frac { p }{ 0 } } $ (because $(\frac{n - 1}{n})^p$ tends to 1).

But as I understood here, limit of this sequence is $\frac{1}{2}$.

Where did I go wrong in my reasoning and why is my answer not correct?

Someone
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2 Answers2

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The final identity $$ \lim_{n\to \infty }{ \frac { ( p+1 )-n+(n-1)(\frac{n - 1}{n})^p }{ ( p+1 ) ( 1-{ (\frac{n-1}{n}) }^{ p } ) } } = \lim _{n\to \infty }{ \frac { p }{ 0 } } $$ is wrong. It is correct that $(\frac{n - 1}{n})^p$ tends to one, but that does not imply that $1 - n + (n-1)(\frac{n - 1}{n})^p$ tends to zero and and can be ignored in the numerator. Also the denominator tends to $p+1$, not to zero.

For a correct solution using the Stolz–Cesàro theorem see for example haqnatural's answer here.

Martin R
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$$\lim _{ n\rightarrow \infty }{ \frac { ( p+1 ) { n }^{ p }-{n}^{ p+1 }+{ (n-1) }^{ p+1 } }{ ( p+1 ) ( { n }^{ p }-{ (n-1) }^{ p } ) } } $$ set $n=\frac{1}{x}$ the limit becomes $$\lim _{ x\rightarrow 0 }\frac{(p+1) \left(\frac{1}{x}\right)^p-\left(\frac{1}{x}\right)^{p+1}+\left(\frac{1}{x}-1\right)^{p+1}}{(p+1) \left(\left(\frac{1}{x}\right)^p-\left(\frac{1}{x}-1\right)^p\right)}$$ dramatic simplification $$\lim _{ x\rightarrow 0 }\frac{\frac{p}{1-(1-x)^p}+1-\frac{1}{x}}{p+1}$$ MacLaurin expansion $$\frac{\frac{p}{1-(1-x)^p}+1-\frac{1}{x}}{p+1}\sim \frac{1}{2}+\frac{1}{12} (p-1) x+O\left(x^2\right)$$ as $x\to 0$ we have $$\frac{\frac{p}{1-(1-x)^p}+1-\frac{1}{x}}{p+1}\to \frac{1}{2}$$

Raffaele
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