Calculus Problem

[jwill22]

Find the limit of the sequence if it converges;otherwise indicate divergence.
a[sub:13urg49k]n[/sub:13urg49k] = ln(7n-6) - ln(5n+7)

The sequence i got starting with a[sub:13urg49k]5[/sub:13urg49k] was -.098,-.02,.02,.06,.09,.11,.13,.15,.16

So i think it converges, but I'm not sure to what. Can somebody help with this problem

 

[galactus]

Note that \(\displaystyle ln(7n-6)-ln(5n+7)=ln\left(\frac{7n-6}{5n+7}\right)=ln\left(\frac{7}{5}-\frac{79}{5(5n+7)}\right)\)

As n grows larger and larger, the n term tends toward 0 and all that remains is the 7/5. See what the sequence converges to now?.

 

[lookagain]

Find the limit of the sequence if it converges;otherwise indicate divergence.
a[sub:sftwk40t]n[/sub:sftwk40t] = ln(7n-6) - ln(5n+7)

The sequence i got starting with a[sub:sftwk40t]5[/sub:sftwk40t] was -.098,-.02,.02,.06,.09,.11,.13,.15,.16

So i think it converges, but I'm not sure to what. Can somebody help with this problem

jwill22, you can stay with the method of galactus, if you would prefer to.

Also:

\(\displaystyle \lim_{n \to \infty}\ln({\frac{7n - 6}{5n + 7})\)

Multiply the numerator and the denominator each by \(\displaystyle \frac{1}{n}\):

\(\displaystyle \lim_{n \to \infty} \ln( \frac{7 - \frac{6}{n}}{5 + \frac{7}{n}} ) =\)

\(\displaystyle \ln(\frac{7}{5}) \approx 0.336\)

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jwill22,

I cannot tell from the values in your sequence that this would be convergent.