I tried to find the value of the sum
$$
\sum_{n=1}^{\infty} \frac{1}{2022^{n}} \cdot\left[\frac{2021^{n}}{2022}\right]
$$
where $[x]$ represents the integer part of $x$.
Here some of my thoutghs:
a) It definetly convergs, since the problem directly asks for a value.
b) Representing ${2021^{n}} $ as $(2022-1)^{n}$, might be possible to cancel some terms with a binomial expansion
$$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n – k} b^k = \binom{n}{0} a^n + \binom{n}{1} a^{n-1}b + \dots + \binom{n}{k} a^{n-k}b^k + \dots + \binom{n}{n} b^n.$$
c) Representing the general term of the sum as
$$
\frac{1}{2022^{n}} \cdot\left[\frac{2021}{2022} \cdot\underbrace{(2021)\cdots (2021)}_{n-1}\right]
$$
and then find the limit of it as $n$ approches infinity.
The expresion, inside the square brackets I saw it as: "taking the 99,95% of $2021^{n-1}$".
Since for $n>1$ it is always an improper fraction, our integer part is always different of zero.
I supposed that we could take $\displaystyle \frac{2021}{2022}$ as $1$, and then we have that after taking the integer part
$$\frac{(2022-1)^{n-1}}{2022^n}$$
At this moment I recall my though b) and assumed that we could; use binomial expasion;take the first term $\displaystyle \frac{1}{2022} \approx 0,00049$ ans consider the rest as error that just sum up and approches to zero. I think that the value of the sum equals $0,0005$
What is the answer and how do you solve it in an elegant way?
I got the feeling I helped myself too much with calculator trying to solve it (understand the behavior).
Best Answer
For the integer part we have that $$\sum_{n=1}^{\infty} \frac{1}{2022^{n}} \cdot\left[\frac{2021^{n}}{2022}\right]=\frac{4084440}{4088483}\approx 0.9990111$$ and therefore for the fractional part we find (recall that $\{x\} = x-[x]$), $$\sum_{n=1}^{\infty} \frac{1}{2022^{n}} \cdot\left\{\frac{2021^{n}}{2022}\right\}=\sum_{n=1}^{\infty} \frac{2021^n}{2022^{n+1}}-\sum_{n=1}^{\infty} \frac{1}{2022^{n}} \cdot\left[\frac{2021^{n}}{2022}\right]\\=\frac{2021}{2022}-\frac{4084440}{4088483}=\frac{4086463}{8266912626}\approx 0.0004943.$$
As regards the first sum, we start by using your approach: \begin{align} \sum_{n=1}^{\infty} \frac{1}{2022^{n}} \cdot\left[\frac{2021^{n}}{2022}\right]&=\sum_{n=1}^{\infty} \frac{1}{2022^{n}} \cdot\left[\frac{(2022-1)^{n}}{2022}\right]\\ &=\sum_{n=1}^{\infty} \frac{1}{2022^{n}} \cdot\left(\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}2022^{n-1-k}+\left[\frac{(-1)^n}{2022}\right]\right)\\ &=\sum_{n=1}^{\infty} \frac{1}{2022^{n}} \cdot\left(\frac{2021^n-(-1)^n}{2022}+\frac{(-1)^{n}-1}{2}\right)\\ &=\frac{4084440}{4088483}. \end{align} If we replace $2022$ with an integer $N>1$, the above sum is equal to $\frac{N(N-2)}{N^2-1}$.