Ask Uncle Colin: Why is $e$ not 1?

Dear Uncle Colin,

If $e = \left( 1+ \frac{1}{n} \right)^n$ when $n = \infty$, how come it isn’t 1? Surely $1 + \frac{1}{\infty}$ is just 1?

- I’m Not Finding It Natural, It’s Terribly Yucky

Hi, INFINITY, and thanks for your message.

You have fallen into one of maths’s classic traps: infinity1 is not a number - you can’t just plug it into equations and expect to get sensible things out. (If infinity was a number, we could flip your argument around and say “$1 + \frac{1}{n}$ is a bit more than 1, and if we multiply infinitely many of those together, it gets bigger each time, so it must go to infinity.” It doesn’t work that way, either.)

Instead, we need to think about limits: What happens when $n$ gets really big?

Binomial!

We can expand the expression using the binomial series: $\br{1 + \frac{1}{n}}^n = 1 + n \times \frac{1}{n} + \frac{n(n-1)}{2} \frac{1}{n^2} + \dots$

That works out to be $1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots$, minus a load of terms with $n$s on the bottom.

But when $n$ gets big, those all get extremely small, leaving you with $1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots = e$.

I hope that helps!

- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. hey! that’s your name! []

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