به انگلیسی

Adding infinitely many terms

Mathematics · Calculus & Analysis · ages 19-20

Name ______________________ Date ____________

What is the sum of the geometric series 6 + 2 + 2/3 + 2/9 and so on?
Answer: ______________
What is the sum of the geometric series 1 + 1/2 + 1/4 + ...?
Answer: ______________
The sum of a series is defined as the limit of its partial sums, when that limit exists.
Circle one: True False
An infinite geometric series has a nonzero first term and a real common ratio. Which condition is necessary and sufficient for it to have a finite sum?
- The ratio has absolute value below 1
- The ratio is positive and less than 2
- The first term has absolute value below 1
The series with terms 1 over the square root of n has terms tending to zero. What does the term test say about it?
- It converges because the terms tend to zero
- It diverges because the terms tend to zero
- The test gives no conclusion
What is the sum of the geometric series 4 + 4/3 + 4/9 + ...?
Answer: ______________
For n = 1, 2, 3, ..., a series has terms 1/[n(n + 1)] and partial sums S_N = N/(N + 1). What is its sum?
Answer: ______________
Starting at n = 1, the terms 2/[n(n + 2)] split as 1/n - 1/(n + 2). The Nth partial sum is 3/2 - 1/(N + 1) - 1/(N + 2). What is the sum of the series?
Answer: ______________
Which of these series must diverge?
- One whose terms shrink to zero and decrease
- One whose partial sums stay below 10
- One whose nth term tends to 5
Which series is shown to diverge by the term test alone?
- Terms n/(n + 1)
- Terms 1/n
- Terms 1/(n^2)
- Terms 1/(2^n)

Answer key

For grown-ups. Fold this page away before handing over the rest.

9 · First term 6 with ratio 1/3 gives 6 over 2/3, which is 9.
2 · The first term is 1 and the ratio is 1/2. Since |r| < 1, the sum is 1/(1 - 1/2) = 1/(1/2) = 2.
True · Add finitely many terms first, then let the count grow. When those totals approach a limit, that limit is the sum.
The ratio has absolute value below 1 · With a nonzero first term, an infinite geometric series converges exactly when |r| < 1. Its partial sums then tend to a/(1 - r).
The test gives no conclusion · The term test can only show divergence when terms do not tend to zero. Here it gives no verdict.
6 · The first term is 4 and the ratio is 1/3. The sum is 4/(1 - 1/3) = 4/(2/3) = 6.
1 · The split 1/[n(n + 1)] = 1/n - 1/(n + 1) gives S_N = 1 - 1/(N + 1). The leftover fraction tends to 0, so the sum is 1.
1.5 · Both leftover fractions, 1/(N + 1) and 1/(N + 2), tend to 0. The partial sums therefore tend to 3/2 = 1.5.
One whose nth term tends to 5 · Terms that do not tend to zero violate a necessary condition for convergence.
Terms n/(n + 1) · n/(n + 1) tends to 1, not 0, so the term test proves divergence. The other terms tend to 0. The series with terms 1/n also diverges, but this test cannot show it.