Question

#Challenge_math
Find the mistake in the below proof of 1 = 0.
$\boxed{\begin{aligned} 1 & = \lim_{n\to \infty} 1\\&= \lim_{n\to \infty} \frac{n}{n}\\&=\lim_{n\to \infty} \bigg\{\underbrace{\frac{1}{n} + \frac{1}{n} + \frac{1}{n} + ...+ \frac{1}{n} }_{\sf n\ terms}\bigg\}\\& = \underbrace{\lim_{n\to \infty} \frac1n + \lim_{n\to \infty} \frac 1 n + \lim_{n\to \infty} \frac 1n + ... + \lim_{n\to \infty} \frac 1n}_{\sf n \ terms} \\& =\underbrace{ 0 + 0+0+...+0}_\sf{n\ terms}\\&= 0\end{aligned}}$

$\bullet$ No spam else will be reported.
$\bullet$ Answer with enough explanation will be marked as brainliest.

Answer 1

$\large\underline{\sf{Solution-}}$

Given proof for 0 = 1

$\rm \: 1 = \displaystyle\lim_{n\to \infty} 1$

$\rm \: = \displaystyle\lim_{n\to \infty} \frac{n}{n}$

$\\=\lim_{n\to \infty} \bigg\{\underbrace{\frac{1}{n} + \frac{1}{n} + \frac{1}{n} + ...+ \frac{1}{n} }_{\sf n\ terms}\bigg\}\\ = \underbrace{\lim_{n\to \infty} \frac1n + \lim_{n\to \infty} \frac 1 n + \lim_{n\to \infty} \frac 1n + ... + \lim_{n\to \infty} \frac 1n}_{\sf n \ terms}$

$\\ =\underbrace{ 0 + 0+0+...+0} \\ \sf{n\ terms}\\= 0$

Now, the fallacy of this question is the very first step.

We know that,

$\rm \: \displaystyle\sum_{k=1}^{n} \: 1 \: = \: n$

So,

$\rm \: \displaystyle\lim_{n\to \infty}\displaystyle\sum_{k=1}^{n} \: 1 \: \ne \: n \:$

So, the following step is invalid

As n is infinite, we can't take as finite number of terms.

$\\=\lim_{n\to \infty} \bigg\{\underbrace{\frac{1}{n} + \frac{1}{n} + \frac{1}{n} + ...+ \frac{1}{n} }_{\sf n\ terms}\bigg\}\\ = \underbrace{\lim_{n\to \infty} \frac1n + \lim_{n\to \infty} \frac 1 n + \lim_{n\to \infty} \frac 1n + ... + \lim_{n\to \infty} \frac 1n}_{\sf n \ terms}$

Hence, the above result is not valid.

$\rm\implies \:0 \: \ne \: 1$

$\rule{190pt}{2pt}$

Additional Information :-

Some important summations :-

$\boxed{\rm{  \:\displaystyle\sum_{k=1}^{n} k \: = \: \frac{n(n + 1)}{2} \: }}$

$\boxed{\rm{  \:\displaystyle\sum_{k=1}^{n} {k}^{2} \: = \: \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\boxed{\rm{  \:\displaystyle\sum_{k=1}^{n} {k}^{3} \: = \: \bigg[\frac{n(n + 1)}{2}\bigg]^{2} \: }}$