Final answer:
The first series converges using the ratio test with a ratio of (2/3), indicating convergence. The second series is a telescoping series and converges to 1, since most terms cancel out except for the first term from the numerator.
Step-by-step explanation:
To find the limits of the given series, we need to determine whether each series converges or diverges.
Series (a)
The first series is Σ_(n=1)^[∞] (5 × 2^(n-1)) / (3^(n+1)). To test for convergence, we can use the ratio test, where we take the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term. If the limit of the ratio is less than one, the series converges; if it is greater than one or equals to one, the series diverges.
Using the ratio test:
- Find the (n+1)th term: (5 × 2^n)/(3^(n+2)).
- Find the ratio of the (n+1)th to nth term and simplify.
- Take the limit as n approaches infinity of the absolute value of this ratio.
The ratio is (2/3) which is less than one, so the series converges.
Series (b)
The second series is Σ_(n=1)^[∞] (1/n - 1/(n+2)). This is a telescoping series. Telescoping series are series where many terms cancel out when you add them all up. To find the limit, list out the first few terms and observe how they cancel:
- 1 - 1/3
- + 1/2 - 1/4
- + 1/3 - 1/5
- + 1/4 - 1/6, etc.
Most terms cancel, and we're left with a few terms that do not cancel. The sum approaches the sum of these left-over terms as n approaches infinity.
The limit for the series (b) is 1 because the negative terms cancel out leaving the first term from the numerator.