Which of the following series converge?
A. 1 + 1/2^2 + 1/3^2 + ... + 1/n^2 + ...
B. 1 + 1/2 + 1/3 + ... + 1/n + ...
C. 1 - 1/3 + 1/3^2 - ... + ((-1)^(n+1))/(3^(n-1)) + ...
I think A and C are the answer.
A does, since ∑1/n^p converges for p>1
B does not -- the Harmonic Series
C does, since ∑1/3^n is just a geometric series with r<1
you are correct
Well, let me clown around and check that for you!
For series A, we have 1 + 1/2^2 + 1/3^2 + ... + 1/n^2 + ..., which is known as the "Basel problem." The great mathematician Euler found that this series indeed converges, and its sum is π^2/6. So, you are absolutely right about A!
As for series C, we have 1 - 1/3 + 1/3^2 - ... + ((-1)^(n+1))/(3^(n-1)) + ..., also known as the "Leibniz series." This series is an alternating series, meaning each term has alternating signs. It is indeed convergent, but its sum isn't quite as friendly as π^2/6. The sum is actually (3 - √3)/6. So, you are correct once again about C!
But what about series B? Are you going to leave it out like the lonely odd number at a party? Well, don't worry because series B, 1 + 1/2 + 1/3 + ... + 1/n + ..., is the famous harmonic series. Unfortunately, this series is divergent, meaning it goes to infinity. So, B is the odd one out and does not converge. Poor B!
So, your intuition was spot on! You got two out of three, which is still pretty impressive. Keep up the mathematical humor, my friend!
To determine if these series converge, we can apply the appropriate convergence tests.
A. 1 + 1/2^2 + 1/3^2 + ... + 1/n^2 + ...
This is the harmonic series with squared terms. To check convergence, we can use the Integral Test. Let's integrate the function f(x) = 1/x^2 from 1 to infinity.
∫(1/x^2) dx = -1/x | from 1 to infinity
The integral of the function is -1/x. Evaluating the definite integral, we have:
[-1/x] | from 1 to infinity = -1/infinity - (-1/1) = 0 - (-1) = 1
Since the integral is finite (1 in this case), the series converges.
B. 1 + 1/2 + 1/3 + ... + 1/n + ...
This is the harmonic series. To check convergence, we can use the Divergence Test. The Divergence Test states that if the limit of the terms of a series does not approach zero, then the series diverges.
In this case, the limit as n approaches infinity of the terms 1/n does approach zero. Thus, we cannot conclude anything about convergence or divergence using the Divergence Test. However, it's important to note that the harmonic series, in general, diverges.
C. 1 - 1/3 + 1/3^2 - ... + ((-1)^(n+1))/(3^(n-1)) + ...
This is an alternating series. To check convergence or divergence, we can use the Alternating Series Test. For an alternating series to converge, the absolute values of the terms must decrease and approach zero.
Here, the absolute values of the terms (1/3^n) decrease and approach zero as n increases. Therefore, the series converges.
In summary, out of the given series, both A and C converge.
To determine whether a series converges or diverges, we need to apply some convergence tests. Let's analyze each series one by one:
A. 1 + 1/2^2 + 1/3^2 + ... + 1/n^2 + ...
This is a series of terms that decrease in value. This series is known as the "p-series" with the exponent p = 2. The series 1/n^p converges if p > 1 and diverges if p <= 1. In this case, p = 2, so the series converges. This means option A converges.
B. 1 + 1/2 + 1/3 + ... + 1/n + ...
This series is known as the "harmonic series." Unfortunately, the harmonic series diverges. This is a well-known mathematical result. This means option B diverges.
C. 1 - 1/3 + 1/3^2 - ... + ((-1)^(n+1))/(3^(n-1)) + ...
This is an alternating series because the signs alternate from positive to negative. To check convergence for an alternating series, we need to verify two conditions:
1. The absolute value of the terms must converge to zero. In this instance, as n increases, ((-1)^(n+1))/(3^(n-1)) approaches zero, so the first condition is met.
2. The terms must be decreasing in absolute value. In this case, the terms are decreasing as well since ((-1)^(n+1))/(3^(n-1)) < ((-1)^n)/(3^(n-2)) for all positive integers n.
Both conditions are satisfied, which means the series converges. This confirms that option C is also correct.
Therefore, both options A and C converge.