O limita

[Marius Dragoi]

Sa se calculeze limita:

\( \lim_{n\to\infty} n^{p-pq-1} \sum_{k=0}^{n-1}{(({k+1})^q - k^q)}^p \),

unde p si q sunt doua numere naturale pozitive.

Boros Francisc

[Virgil Nicula]

Sa se calculeze limita \( \lim_{n\to\infty} n^{p-pq-1} \sum_{k=0}^{n-1}{(({k+1})^q - k^q)}^p \) unde \( p>0 \) si \( q>0 \) (nu neaparat naturale !).

Demonstratie.

\( \lim_{n\to\infty} n^{p-pq-1}\cdot \sum_{k=0}^{n-1}\left[({k+1})^q - k^q\right]^p=\lim_{n\to\infty}\frac {\sum_{k=0}^{n-1}\left[({k+1})^q - k^q\right]^p}{n^{1+p(q-1)}}\ \stackrel{ST.}{=}\ \lim_{n\to\infty}\frac {\left[({n+1})^q - n^q\right]^p}{(n+1)^{1+p(q-1)}-n^{1+p(q-1)}}= \)

\( \lim_{n\to\infty}\left[\frac {\left(\frac {n+1}{n}\right)^q-1}{\frac {n+1}{n}-1}\right]^p\cdot\frac {n^{p(q-1)}}{(n+1)^{1+p(q-1)}-n^{1+p(q-1)}}=\frac {q^p}{p(q-1)+1}\Longrightarrow\overline{\underline{\left\|\ \lim_{n\to\infty} n^{p-pq-1}\cdot \sum_{k=0}^{n-1}\left[({k+1})^q - k^q\right]^p=\frac {q^p}{p(q-1)+1}\ \right\|}} \) .

Observatie. \( \ \frac {n^{p(q-1)}}{(n+1)^{1+p(q-1)}-n^{1+p(q-1)}}=\frac {1}{(n+1)\left(1+\frac 1n\right)^{p(q-1)}-n}=\frac {1}{n\left[\left(1+\frac 1n\right)^{p(q-1)}-1\right]+\left(1+\frac 1n\right)^{p(q-1)}}= \)

\( \frac {1}{\frac {\left(1+\frac 1n\right)^{p(q-1)}-1}{\left(1+\frac 1n\right)-1}+\left(1+\frac 1n\right)^{p(q-1)}} \) . Asadar, \( \lim_{n\to\infty}\ \frac {n^{p(q-1)}}{(n+1)^{1+p(q-1)}-n^{1+p(q-1)}}=\lim_{n\to\infty}\ \frac {1}{\frac {\left(1+\frac 1n\right)^{p(q-1)}-1}{\left(1+\frac 1n\right)-1}+\left(1+\frac 1n\right)^{p(q-1)}}=\frac {1}{p(q-1)+1} \) .

Am folosit doar limita remarcabila \( \left\|\ \begin{array}{c}
a_n\rightarrow 1\\\\
a_n\ne 1\ ,\ \left(\forall\right)\ n\in \mathbb N\end{array}\ \right\|\ \Longrightarrow\ \frac {a_n^r-1}{a_n-1}=r\ ,\ \left(\forall\right)\ r\in\mathbb R \) .

Comentariu.

Limita este frumoasa, insa nu depaseste nivelul fazei locale O.M. Observ ca problema are autorul mentionat.
Imi poti spune Marius, te rog, daca a aparut intr-o carte/culegere sau o revista de matematica ?! Multumesc !

[Marius Dragoi]

Am fost contactat de catre autor personal si am pus aceasta problema ajutand-ul pe domnul profesor din cauza faptului ca dansul nu se pricepe momentan cu limbajul \( LaTeX \).

[Laurentiu Tucaa]

Problema e la nivelul clasei a 11-a,nu inteleg de ce a fost pusa la a12-a.