I am so confused that I cannot figure out when to use divergence test and when to use limit comparison test.
For Example: $\sum _{ k=0 }^{ \infty }{ \frac { 4k+7 }{ 11k^ 2+17 } } $
I solved it by $\sum _{ k=0 }^{ \infty }{ \frac { 4k+7 }{ 11k^ 2+17 } } $$\le $$\sum _{ k=0 }^{ \infty }{ \frac { 1 }{ k } } $
and then used p- series test.
But, we can also use divergence test...
Which is the best approach and when to use what
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$\begingroup$The divergence test tells us that if the limit of the summand (the term in the summation) is not zero, then the infinite series must diverge. However, the divergence test does not tell us anything about the series in question if the limit is $0$.
So in the example you gave, since $$lim_{k\rightarrow \infty} \frac{4k+7}{11k^2+17} = 0$$ the divergence test actually doesn't help us in solving whether or not this series diverges or converges.
So we need to try something else, and you're on the right track for using the comparison test, but since the series $$\sum_{k=1}^{\infty} \frac{1}{k}$$ diverges, we want to show that our summation is larger than the series $\sum_{k=1}^{\infty} \frac{1}{k}$, not smaller. (Knowing a sum is smaller than or equal to infinity doesn't really tell us anything about the series.)
All is not lost though! Since $$\frac{4}{11k} < \frac{4k+7}{11k^2+17}$$, and since $$\frac{4}{11} \sum_{k=1}^{\infty} \frac{1}{k}$$ diverges by the p-test, we can now conclude by the comparison test that $\sum_{k=1}^{\infty}\frac{4k+7}{11k^2+17}$ diverges.
So just remember that we can only use the divergence test when the limit of the summand does not go to zero, and that tells us immediately that the series diverges.
$\endgroup$ $\begingroup$The OP's solution is incorrect. Proving that a series is less than a known divergent series (using comparison test), does not allow ANY conclusion about the series.
Instead, I recommend the Limit Comparison Test instead of the comparison test, also with $\sum \frac{1}{n}$. In this case, it will be conclusive, and will prove divergence.
$\endgroup$ 2 $\begingroup$In my somewhat limited experience, I'd say as a rule of thumb use the Limit Comparison Test when its clear a series you know is divergent is related to that series by having the same overall order.
E.g $\frac{1}{k}\rightarrow\frac{Order 0}{Order 1}\rightarrow$ Overall Order $=-1$.
While $\frac{4k+7}{11k^2+17}\rightarrow\frac{Order1}{Order2}\rightarrow$ Overall order $=-1$
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