# `zinCC` and for any real `epsi>0` it exists a `NinNN` : `sum_(k=N+1)^oo [|z|^k/(k!)] < epsi/3.0`Then also: `sum_(k=N+1) ^oo [ ((n),(k)) (z)^k /n^k ]< epsi/3.0` for all natural N

### 1 Answer | Add Yours

First question :

`AA zinCC, sum_(k=0)^oo |z|^k/(k!)=e^|z| `

Therefore, `AA zinCC,`````` sum_(k=N+1)^oo |z|^k/(k!) ` is the rest of a convergent sum.

Hence `AAzinCC, AAepsi>0EEN>0` such that `sum_(k=N+1)^oo |z|^k/(k!) <epsi`

if it is true for any `epsi`, it is particularly true for `epsi/3`

**Hence **`AAzinCC, AAepsi>0EEN>0`** such that` `**` `** **`sum_(k=N+1)^oo |z|^k/(k!) <epsi/3`

**Second question :**

`( ()^n _k) z^k/n^k=(n(n-1)(n-2)...(n-k+1))/n^k z^k/(k!)`

`(()^n _k)z^k/n^k=(n/n)((n-1)/n)((n-2)/n)...((n-k+1)/n) z^k/(k!)`

`|(()^n _k)z^k/n^k|<=1.1. 1... 1 |z|^k/(k!)` because each fraction `(n-?)/n <=1`

If we sum the previous inequalities, we obtain

` ``sum_(k=N+1)^oo|(()^n _k)z^k/n^k|<=sum_(k=N+1)^oo|z|^k/(k!) `

From question 1 and using triangular inequalities for complex numbers.

**Conclusion**

`AAzinCC, AAepsi>0EEN>0`

** **` `** **

` `

` `` ``|sum_(k=N+1)^oo(()^n _k)z^k/n^k|<=``sum_(k=N+1)^oo|(()^n _k)z^k/n^k|``<=sum_(k=N+1)^oo|z|^k/(k!)<epsi/3`` `

``` `

` `

**` `**