First, suppose that `a gt 0 , ` then `b = 1 / a gt 1 , ` and `b = 1 + c , ` where `c gt 0 .`

The expression `n a ^ n ` may be rewritten as `n / b ^ n = n / ( 1 + c ) ^ n .`

Now recall the binomial theorem. It states that `( x + y ) ^ n = x ^ n + n x + ( n ( n - 1 ) ) / 2 x ^ 2 + ... ` (because `n -> oo , ` we can consider `n gt= 2 ` only). Note that for `x , y gt= 0 ` all summands are nonnegative.

Because of this, `( 1 + c ) ^ n gt= 1 + n c + ( n ( n - 1 ) ) / 2 c ^ 2 gt= ( n ( n - 1 ) ) / 2 c ^ 2 . ` Further, this gives us that

`0 <= n a ^ n lt= n / ( ( n ( n - 1 ) ) / 2 c ^ 2 ) = 2 / c ^ 2 1 / ( n - 1 ) ,`

where the right side clearly tends to zero as `n ` tends to `oo . ` We can use squeeze theorem now and conclude that `n a ^n -> 0 .`

If `a ` is negative, then note that `|na^n| = n|a|^n -> 0 , ` so `na^n->0, ` too.

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