23–38. Divergence, Integral, and p-series Tests Use the Diverg...

Question

23–38. Divergence, Integral, and p-series Tests Use the Divergence Test, the Integral Test, or the p-series test to determine whether the following series converge.

∑ (k = 1 to ∞) (k / (k + 10))ᵏ

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Test the convergence of the series, the sum evaluated at M equals 1 to positive infinity of parenthesis M divided by M + 7 to the power of M. Awesome. So it appears for this particular prom we're asked to test the convergence of this particular series that is provided to us by the prom itself. So with that in mind, our final answer will be either a converges or B diverges. So once again we're ultimately trying to test the convergence of this particular series that is provided to us by the problem itself. So in order to do that, our first step that we need to take is we need to recall and use the divergence test. And as we should recall, the divergence test states that if The limit as M approaches positive infinity of a subscript M cannot equal 0 or it does not exist, then the series, the sum. Of a subscript M diverges. For this particular case, a subscript M, which I'll just call AM for short, AM is equal to parentheses M divided by M + 7 in parentheses to the power of M. Fantastic. Our second step now is we need to compute the limit of the terms, so we need to take the limit as M approaches positive infinity. So the limit as M approaches positive infinity of M divided by M + 7 to the power of M means it's M plus. So once again we have M divided by M + 7 in parenthesis to the power of M. And we need to rewrite this to state that M divided by M + 7 is equal to 1 minus 7 divided by M + 7. So we can go ahead and write that AM is equal to parentheses 1 minus 7 divided by m + 7 to the power of Ms AM will be equal to parentheses of 1 minus 7 divided by m + 7 once again to the power of M. Fantastic. So now we need to take the natural logarithm. So we need to take the natural log of AM is equal to M multiplied by, so it's gonna be M multiplied by the natural log of 1 minus 7 divided by M + 7. Fantastic. So now we need to use approximation. So the natural log of 1 minus X will be approximately equal to negative X for small values of X, meaning that the natural log of AM will be, so once again this natural log of AM will be approximately equal to M multiplied by negative 7 divided by M plus 7. Fantastic. So moving right along here, we now need to simplify, so we need to take the natural log of Am is approximately equal to -7M divided by m + 7. So we need to note that as M approaches positive infinity, that will mean that 7 M divided by M + 7 will approach 7. So that will mean that the natural log of AM will approach -7. Fantastic. So moving right along here. So therefore, AM will approach E to the power of -7. As M approaches positive infinity. Fantastic. Our 3rd step now is we need to recall and apply the divergence test. For this particular prompt, so when you take the limit, as M approaches positive infinity of A, subscript M is equal to E the power of -7 cannot equal 0. So therefore by the divergence test, because the series cannot equal 0, the series will diverge. So because we perform the divergence test and based on the divergence test, the series diverges. So let's make a note for D in quotation marks for it diverges, and that is our final answer. So looking at our multiple choice answers, our final answer, without a doubt has to be multiple choice answer letter B, which once again is diverges. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

23–38. Divergence, Integral, and p-series Tests Use the Divergence Test, the Integral Test, or the p-series test to determine whether the following series converge.
∑ (k = 1 to ∞) (k / (k + 10))ᵏ

Key Concepts

Divergence Test

Integral Test

p-Series Test

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