# Convergence Test Help

#### Senn

##### New member

Determine if this series diverges, converges conditionally, or converges absolutely.

I said the series converges conditionally due to AST because the limit of this series when positive is 0, but the positive series is diverging when compared to 1/k using the limit comparison test.
However, the answer is that this series is diverging, and I wanted to know what I am doing wrong.

#### Dr.Peterson

##### Elite Member
View attachment 33596
Determine if this series diverges, converges conditionally, or converges absolutely.

I said the series converges conditionally due to AST because the limit of this series when positive is 0, but the positive series is diverging when compared to 1/k using the limit comparison test.
However, the answer is that this series is diverging, and I wanted to know what I am doing wrong.
Who says it diverges? Do you fully trust that source? Or is your own work sufficiently convincing to consider that maybe the source is wrong? (Or that you misread it?)

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#### blamocur

##### Senior Member
View attachment 33596
Determine if this series diverges, converges conditionally, or converges absolutely.

I said the series converges conditionally due to AST because the limit of this series when positive is 0, but the positive series is diverging when compared to 1/k using the limit comparison test.
However, the answer is that this series is diverging, and I wanted to know what I am doing wrong.
You answer looks correct to me, and my simple script agrees: the series converge to approximately -0.28094 (with [imath]\infty[/imath] set to [imath]5\times10^8[/imath] )

#### lookagain

##### Elite Member
. . . the series converge to approximately -0.28094 (with [imath]\infty[/imath] set to [imath]5\times10^8[/imath] )

WolframAlpha shows that it approximately equals 0.396454 at this link:

#### blamocur

##### Senior Member
I did get my sign wrong, i.e. it should 0.28094. As for WalframAlpha, strangely it returns 0.347280... when I asked for more digits.

#### blamocur

##### Senior Member
I did get my sign wrong, i.e. it should 0.28094. As for WalframAlpha, strangely it returns 0.347280... when I asked for more digits.
... and WalframAlpha agrees with my answer when I ask for [imath]\sum_{k=2}^\infty (-1)^k 9 \frac{\log (k)}{k+6}[/imath] :

#### topsquark

##### Senior Member
I did get my sign wrong, i.e. it should 0.28094. As for WalframAlpha, strangely it returns 0.347280... when I asked for more digits.
To the corner with you!

-Dan

#### Subhotosh Khan

##### Super Moderator
Staff member
... and WalframAlpha agrees with my answer when I ask for [imath]\sum_{k=2}^\infty (-1)^k 9 \frac{\log (k)}{k+6}[/imath] :
Where did that '9' come from?

#### BigBeachBanana

##### Senior Member
Where did that '9' come from?
$$\displaystyle \log(k^9)=9\log(k)$$

#### BigBeachBanana

##### Senior Member
My script agrees with blamocur.

I find lookagain's link odd as it includes [imath]i?[/imath]

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#### nasi112

##### Full Member
View attachment 33596
Determine if this series diverges, converges conditionally, or converges absolutely.

I said the series converges conditionally due to AST because the limit of this series when positive is 0, but the positive series is diverging when compared to 1/k using the limit comparison test.
However, the answer is that this series is diverging, and I wanted to know what I am doing wrong.
Always trust the tests.

$$\displaystyle \frac{9}{k+6} < \frac{9\ln k}{k + 6}$$

$$\displaystyle \sum_{k=2}^{\infty}\frac{9}{k+6}$$ is a divergent p-series, so $$\displaystyle \sum_{k=2}^{\infty}\frac{\ln(k^9)}{k+6}$$, diverges by the comparison test.

We have two options now, it's whether $$\displaystyle \sum_{k=2}^{\infty}(-1)^k\frac{\ln(k^9)}{k+6}$$ diverges or converges conditionally.

Now, I will try the alternating series test.

$$\displaystyle \lim_{k \rightarrow \infty}\frac{9\ln(k)}{k+6} = 0$$

$$\displaystyle \frac{9\ln(k+1)}{(k+7)} \times \frac{(k+6)}{9\ln(k)} < 1$$ for $$\displaystyle k \geq 7$$

Therefore, $$\displaystyle \sum_{k=2}^{\infty}(-1)^k\frac{\ln(k^9)}{k+6}$$, converges conditionally.

Note that I started the comparison test first. A better practice is to start the alternating series first.

If $$\displaystyle \frac{9\ln(k+1)}{(k+7)} \times \frac{(k+6)}{9\ln(k)}$$ is not obvious whether it is $$\displaystyle > 1$$ or $$\displaystyle < 1$$, take the limit.

$$\displaystyle \lim_{k \rightarrow \infty}\frac{9\ln(k+1)}{(k+7)} \times \frac{(k+6)}{9\ln(k)} = 1$$

If the limit is $$\displaystyle 1$$, here is the trick,

You know that $$\displaystyle \frac{\ln(k + 1)}{\ln(k)}$$ is always $$\displaystyle > 1$$. If its limit is $$\displaystyle 1$$, it means that it approaches $$\displaystyle 1$$ from above. In other words, $$\displaystyle \frac{\ln(k + 1)}{\ln(k)} \approx 1.0000001$$.

You know that $$\displaystyle \frac{k + 6}{k+7}$$ is always $$\displaystyle < 1$$. If its limit is $$\displaystyle 1$$, it means that it approaches $$\displaystyle 1$$ from below. In other words, $$\displaystyle \frac{k + 6}{k + 7} \approx 0.9999999$$.

And $$\displaystyle 1.0000001 \times 0.9999999$$ must be $$\displaystyle < 1$$

#### blamocur

##### Senior Member
Always trust the tests.

$$\displaystyle \frac{9}{k+6} < \frac{9\ln k}{k + 6}$$

$$\displaystyle \sum_{k=2}^{\infty}\frac{9}{k+6}$$ is a divergent p-series, so $$\displaystyle \sum_{k=2}^{\infty}\frac{\ln(k^9)}{k+6}$$, diverges by the comparison test.

We have two options now, it's whether $$\displaystyle \sum_{k=2}^{\infty}(-1)^k\frac{\ln(k^9)}{k+6}$$ diverges or converges conditionally.

Now, I will try the alternating series test.

$$\displaystyle \lim_{k \rightarrow \infty}\frac{9\ln(k)}{k+6} = 0$$

$$\displaystyle \frac{9\ln(k+1)}{(k+7)} \times \frac{(k+6)}{9\ln(k)} < 1$$ for $$\displaystyle k \geq 7$$

Therefore, $$\displaystyle \sum_{k=2}^{\infty}(-1)^k\frac{\ln(k^9)}{k+6}$$, converges conditionally.

Note that I started the comparison test first. A better practice is to start the alternating series first.

If $$\displaystyle \frac{9\ln(k+1)}{(k+7)} \times \frac{(k+6)}{9\ln(k)}$$ is not obvious whether it is $$\displaystyle > 1$$ or $$\displaystyle < 1$$, take the limit.

$$\displaystyle \lim_{k \rightarrow \infty}\frac{9\ln(k+1)}{(k+7)} \times \frac{(k+6)}{9\ln(k)} = 1$$

If the limit is $$\displaystyle 1$$, here is the trick,

You know that $$\displaystyle \frac{\ln(k + 1)}{\ln(k)}$$ is always $$\displaystyle > 1$$. If its limit is $$\displaystyle 1$$, it means that it approaches $$\displaystyle 1$$ from above. In other words, $$\displaystyle \frac{\ln(k + 1)}{\ln(k)} \approx 1.0000001$$.

You know that $$\displaystyle \frac{k + 6}{k+7}$$ is always $$\displaystyle < 1$$. If its limit is $$\displaystyle 1$$, it means that it approaches $$\displaystyle 1$$ from below. In other words, $$\displaystyle \frac{k + 6}{k + 7} \approx 0.9999999$$.

And $$\displaystyle 1.0000001 \times 0.9999999$$ must be $$\displaystyle < 1$$
Where do 1.0000001 and 0.9999999 come from?

#### nasi112

##### Full Member
Where do 1.0000001 and 0.9999999 come from?
Plug a large number and you will have an idea (a pattern) what is going on in $$\displaystyle \infty$$. Say $$\displaystyle 9999999$$.

$$\displaystyle \frac{\ln(9999999 + 1)}{\ln(9999999)} \approx 1.000000006 \approx 1.00000001$$

$$\displaystyle \frac{(9999999 + 6)}{(9999999 + 7)} \approx 0.9999999$$

I should have written $$\displaystyle 1.00000001$$ (7 zeros) instead of $$\displaystyle 1.0000001$$ (6 zeros). But it does not matter much as we are here to compare:

$$\displaystyle 0.99999990$$
$$\displaystyle 1.00000001$$ (We care only for the last digit here $$\displaystyle \geq$$ 1)

$$\displaystyle 1.00000001 \times 0.99999990 < 1$$

If the numbers were, for example,

$$\displaystyle 0.99999999$$
$$\displaystyle 1.00000001$$

As long as the last digit is $$\displaystyle 1$$ or $$\displaystyle 0$$,

$$\displaystyle 1.00000001 \times 0.99999999 < 1$$

If the last digit is not $$\displaystyle 1$$, we look at the digit above it. If it is $$\displaystyle 9$$, then we have an expression $$\displaystyle > 1$$. If the last digit is not $$\displaystyle 1$$ and the digit above it is $$\displaystyle 0$$, then we have an expression $$\displaystyle < 1$$.

These are the most frequent cases that you will encounter:

a. ($$\displaystyle < 1$$)
0.99999999
1.00000001

b. ($$\displaystyle > 1$$)
0.99999999
1.00000002 (or any number above 1)

c. ($$\displaystyle < 1$$)
0.99999990
1.00000001 (or any number above 1)

The position of the digit $$\displaystyle 1$$ is not the last. To understand this, I will give an example below.
d. ($$\displaystyle > 1$$)
0.99999999
1.00010000 (or any number above 1)

-----
Let us flip the series upside down and see what happens.

$$\displaystyle \frac{\ln(9999999)}{\ln(9999999+1)} \approx 0.99999999$$

$$\displaystyle \frac{(9999999 + 7)}{(9999999 + 6)} \approx 1.0000001$$

We are comparing:

0.99999999
1.00000010 (the position of the digit $$\displaystyle 1$$ is not the last)

We have an expression $$\displaystyle > 1$$.

#### Steven G

##### Elite Member
Always trust the tests.

$$\displaystyle \frac{9}{k+6} < \frac{9\ln k}{k + 6}$$
That inequality is not always true (try k=2). You need to say that it is eventually true.

#### Steven G

##### Elite Member
$$\displaystyle \frac{9\ln(k+1)}{(k+7)} \times \frac{(k+6)}{9\ln(k)} < 1$$ for $$\displaystyle k \geq 7$$
And this inequality is true for k≥7 because....
Where is your proof of this??
Saying 1.0000001×0.9999999 must be <1 is true but it is also true that 2x3=6.
What if you had 1.00001*.99999999999 which is greater than 1?
You just can't just say that statements are true just because you want them to be true!

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#### nasi112

##### Full Member
And this inequality is true for k≥7 because....
Where is your proof of this??
Saying 1.0000001×0.9999999 must be <1 is true but it is also true that 2x3=6.
What if you had 1.00001*.99999999999 which is greater than 1?
You just can't just say that statements are true just because you want them to be true!
I cannot prove it by myself. This method was provided by our professor and it is guaranteed to always work. Prove me wrong!

1.00001*.99999999999 > 1

The idea of this method is that we don't really need to know exactly what happens in $$\displaystyle \infty$$. It is enough to understand the behavior of the two functions in the way to infinite.

When plugging a large number (same number) in $$\displaystyle f(x)$$ and $$\displaystyle g(x)$$ and you get $$\displaystyle f(x) \times g(x) < 1$$ or $$\displaystyle f(x) \times g(x) > 1$$, no matter how further you go, you will get the same result.

#### nasi112

##### Full Member
That inequality is not always true (try k=2). You need to say that it is eventually true.
If we do a comparison in infinite series, we don't care about the first few terms. Even without saying eventually, it is understandable.

#### blamocur

##### Senior Member
I cannot prove it by myself. This method was provided by our professor and it is guaranteed to always work. Prove me wrong!
Unlike the presumption of innocence in the rule of law there is no presumption of correctness in math. I.e., a statement is not considered proven correct until proven wrong.