If $\sum\limits_{n=1}^{2026}\tan^{-1}\left(\dfrac{1}{n^2+n+1}\right) = \tan^{-1}\left(1 - \dfrac{1}{x}\right)$, where $x \neq 0$, then $x = $

The general term of the series is given by:

$T_n = \tan^{-1}\left(\dfrac{1}{n^2+n+1}\right)$

This can be rewritten as:

$T_n = \tan^{-1}\left(\dfrac{1}{1 + n(n+1)}\right) = \tan^{-1}\left(\dfrac{(n+1) - n}{1 + (n+1)n}\right)$

Using the formula $\tan^{-1}\left(\dfrac{x-y}{1+xy}\right) = \tan^{-1}(x) - \tan^{-1}(y)$, we get:

$T_n = \tan^{-1}(n+1) - \tan^{-1}(n)$

Summing from $n=1$ to $n=2026$, we obtain a telescoping series:

$\sum\limits_{n=1}^{2026} T_n = (\tan^{-1}(2) - \tan^{-1}(1)) + (\tan^{-1}(3) - \tan^{-1}(2)) + \dots + (\tan^{-1}(2027) - \tan^{-1}(2026))$

$\sum\limits_{n=1}^{2026} T_n = \tan^{-1}(2027) - \tan^{-1}(1)$

Applying the inverse tangent difference formula again:

$\tan^{-1}(2027) - \tan^{-1}(1) = \tan^{-1}\left(\dfrac{2027 - 1}{1 + 2027 \times 1}\right)$

$= \tan^{-1}\left(\dfrac{2026}{2028}\right) = \tan^{-1}\left(\dfrac{1013}{1014}\right)$

We are given that this sum is equal to $\tan^{-1}\left(1 - \dfrac{1}{x}\right)$. Therefore:

$\tan^{-1}\left(\dfrac{1013}{1014}\right) = \tan^{-1}\left(1 - \dfrac{1}{x}\right)$

Equating the arguments:

$\dfrac{1013}{1014} = 1 - \dfrac{1}{x}$

$\dfrac{1}{x} = 1 - \dfrac{1013}{1014} = \dfrac{1}{1014}$

$x = 1014$

Answer: $1014$