Determine convergence or divergence using any method

Determine convergence or divergence using any method, If you guys could tell me what method you used and show your work that would be really helpful.

∞
Σ 1/n!
n=0


∞
Σ 3/5^n
n=0


Thanks.

Hello,

= = = = = = = = = = = = = = = = = = = = = =
The exponential function can be characterized by:
exp(x) = ∑ (n=0→+∞) xⁿ/n!

Hence, if x=1:
exp(1) = ∑ (n=0→+∞) 1ⁿ/n! = ∑ (n=0→+∞) 1/n!

Thus, your first serie converges:
∑ (n=0→+∞) 1/n! = exp(1) = e

= = = = = = = = = = = = = = = = = = = = = =
Let S(n) = 1 + 1/5 + 1/5² + ... + 1/5ⁿ
Then:

S(n) = 1 + 1/5 + 1/5² + ... + 1/5ⁿ
S(n)/5 = 1/5 + 1/5² + 1/5³ + ... + 1/5ⁿ⁺¹
S(n) - S(n)/5 = 1 - 1/5 + 1/5 - 1/5² + ... + 1/5ⁿ - 1/5ⁿ⁺¹
(4/5).S(n) = 1 - 1/5ⁿ⁺¹
S(n) = ¼(5 - 1/5ⁿ)

∑ (n=0→+∞) 1/5ⁿ = Lim (n=0→+∞) S(n)
   = Lim (n=0→+∞) ¼(5 - 1/5ⁿ)
   = ¼ × Lim (n=0→+∞) (5 - 1/5ⁿ)
   = ¼ × 5
   = 5/4

Hence:
∑ (n=0→+∞) 3/5ⁿ = 3 × ∑ (n=0→+∞) 1/5ⁿ = 3 × 5/4 = 15/4

Logically,
Dragon.Jade :-)

For all

∞
Σ 1/n! <=
n=0

∞
Σ 1/n^2
n=1

= pi^2/6.

So the first one converges.

N-1
Σ 3/5^n
n=0

= 3 (1-0.2^N)/(1-0.2)

--> 3/0.8

So the second one also converges.