How do you know if the series #(1/(2n+1))# converges or diverges for (n=1, ∞) ?

Answer 1

#sum_{n=1}^infty 1/{2n+1} = infty#

By comparison, you can say that #2n+1 ~~ n#. They are asymptotically equivalent because
#lim_{n \to \infty} (2n+1)/n = 2#.

So, the series behaves in the same way of

#sum_{n=1}^infty 1/n#,

which is known to be divergent.

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Answer 2

To determine if the series (1/(2n+1)) converges or diverges for n = 1 to infinity, we can use the integral test. We integrate the function f(x) = 1/(2x + 1) from 1 to infinity. If the integral converges, then the series converges; if the integral diverges, then the series also diverges.

∫[1 to ∞] (1/(2x + 1)) dx = [ln|2x + 1|/2] [1 to ∞]

= lim[as t approaches ∞] (ln|2t + 1|/2 - ln|3|/2)

= lim[as t approaches ∞] (ln|2t + 1|/2)

Since the natural logarithm function grows without bound as its argument goes to infinity, the limit of ln|2t + 1|/2 as t approaches infinity diverges.

Therefore, since the integral diverges, the series (1/(2n+1)) also diverges by the integral test.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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