The question breaks down into two separate problems, both of which involve summation from k=0 to infinity (i.e., ∑ from k=0 to infinity). The first sequence is a geometric series, whilst the second appears to be a more complex exponential series.
Let's start by solving the first series:
1) ∑ from k=0 to infinity of (-4/3)^k
This is known as a geometric series because each term is the previous term times a constant. The constant in this case is -4/3 which we will often call r (i.e., r = -4/3).
The general formula to find the sum of an infinite geometric series is S = a / (1 - r), where a is the first term, and r is the common ratio between terms.
In our series, a = 1 because our power, k, starts at 0 (anything raised to power 0 equals 1). So now we plug a and r into our formula. This gives us:
S = 1 / (1 - (-4/3)) = 1 / (1 + 4/3) = 1 / (7/3)
After performing the division, you get approximately 0.4285714285714286.
That's your answer to the first problem.
Now let's discuss the possibility of solving the second series:
2) ∑ from 1 to infinity of 3^(n-1) * e^n
Unfortunately, this series is a bit more complex as it involves raising 3 to the power of (n-1) and multiplying by e^n. In general, the sum of a sequence can be quite difficult to find, as it often depends on various properties of the sequence -- in this case, the relationship between the terms 3^(n-1) and e^n.
In such types of complex series, usually advanced mathematical techniques are used such as calculus or similar methods.
So, even though we can't provide the exact answer to the second problem without these advanced methods, the explanation above provides an understanding of why it's more complex.