Krásný zápis

lim⁡(an±⋅÷bn)=lim⁡an±⋅÷lim⁡bn\lim {\left(a_n \underset{\div}{\underset{\cdot}{\pm}} b_n\right)} = \lim a_n \underset{\div}{\underset{\cdot}{\pm}} \lim b_n

Zajímavý příkládek

Cvičení. Sestrojte posloupnosti (an),(bn),(cn)(a_n), (b_n), (c_n) takové, aby lim⁡n→∞(an+bn)cn≠lim⁡n→∞(an+b)cn\displaystyle \lim_{n\to\infty} {\left(a_n + b_n\right)} c_n \neq \lim_{n\to\infty} {\left(a_n + b\right)} c_n, kde b=lim⁡n→∞bn\displaystyle b = \lim_{n\to\infty} b_n.
Řešenían=1n,bn=1n,cn=na_n = \frac{1}{n}, b_n = \frac{1}{n}, c_n = n

Důkaz součtu čtverců

∑k=1n(k−1)3=∑k=1nk3−3∑k=1nk2+3∑k=1n−∑k=1n1=∑k=0n−1k3=∑k=1nk3−n33∑k=1nk2=n3+3n(n+1)2−n=2n3+3n2+n2=n(n+1)(2n+1)2∑k=1nk2=n(n+1)(2n+1)6\begin{align*}\sum_{k=1}^n (k-1)^3 &= \sum_{k=1}^n k^3 - 3 \sum_{k=1}^n k^2 + 3 \sum_{k=1}^n - \sum_{k=1}^n 1 \\&= \sum_{k=0}^{n-1} k^3 = \sum_{k=1}^n k^3 - n^3 \\ 3 \sum_{k=1}^n k^2 &= n^3 + 3 \frac{n (n+1)}{2} - n \\&= \frac{2 n^3 + 3 n^2 + n}{2} \\&= \frac{n (n+1) (2n+1)}{2} \\ \sum_{k=1}^n k^2 &= \frac{n (n+1) (2n+1)}{6}\end{align*}

Obecně

∑k=1n(km+1−1)=∑l=0m+1(−1)l∑k=1n(m+1l)kl=∑k=0n−1km+1=∑k=1nkm+1−nm+1(m+1)∑k=1nkm=nm+1+∑l=0m−1(−1)l∑k=1n(m+1l)kl∑k=1nkm=nm+1+∑l=0m−1(−1)l∑k=1n(m+1l)klm+1\begin{align*}\sum_{k=1}^n {\left(k^{m+1} - 1\right)} &= \sum_{l=0}^{m+1} (-1)^l \sum_{k=1}^n {m+1 \choose l} k^l \\&= \sum_{k=0}^{n-1} k^{m+1} = \sum_{k=1}^n k^{m+1} - n^{m+1} \\ (m+1) \sum_{k=1}^n k^m &= n^{m+1} + \sum_{l=0}^{m-1} (-1)^l \sum_{k=1}^n {m+1 \choose l} k^l \\ \sum_{k=1}^n k^m &= \frac{n^{m+1} + \sum_{l=0}^{m-1} (-1)^l \sum_{k=1}^n {m+1 \choose l} k^l}{m+1}\end{align*}