`sum_(n=1)^oo 1/(nsqrt(n^2+1))` Use the Limit Comparison Test to determine the convergence or divergence of the series.

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Recall Limit Comparison Test considers two positive series `a_ngt=0` and `b_ngt 0` for all `n` such that the limit from the ratio of two series as:

`lim_(n-gtoo)a_n/b_n=c`

 where `c ` is positive and  finite `( 0ltcltoo)` .

When we satisfy the condition for the limit value,  the two series will have the same properties. Both will either converge or diverges. We may also consider the conditions:

 If  we have  `lim_(n-gtoo)a_n/b_n=0` , we follow: `sum b_n` converges then `sum a_n` converges.

 If  we have `lim_(n-gtoo)a_n/b_n=oo` , we follow: `sum b_n` diverges then `sum a_n` diverges.

 For the given series `sum_ (n=1)^oo 1/(nsqrt(n^2+1))` , we may let `a_n= 1/(nsqrt(n^2+1)).`

Rationalize the denominator:

`1/(nsqrt(n^2+1)) *sqrt(n^2+1)/sqrt(n^2+1) =sqrt(n^2+1)/(n(n^2+1)`

Note: `sqrt(n^2+1)*sqrt(n^2+1) = (sqrt(n^2+1))^2 = n^2+1` .

Ignoring the constants, we get:

`sqrt(n^2+1)/(n(n^2+1)) ~~sqrt(n^2)/(n(n^2)) or 1/n^2`

Note: `sqrt(n^2) =n` . We may cancel it out to simplify.

This gives us a hint that we may apply comparison between the two series: `a_n= 1/(nsqrt(n^2+1))` and  `b_n = 1/n^2` .

 The limit of the ratio of the two series will be:

`lim_(n-gtoo) [1/(nsqrt(n^2+1))]/[1/n^2] =lim_(n-gtoo) 1/(nsqrt(n^2+1))*n^2/1`

                       `=lim_(n-gtoo) n^2/(nsqrt(n^2+1))`

                       ` =lim_(n-gtoo) n/sqrt(n^2+1)`

Apply algebraic techniques to evaluate the limit. We divide by n with the highest exponent which is `n`  or `n^1` . Note: `n` is the same as `sqrt(n^2)`  .

`lim_(n-gtoo) n/(sqrt(n^2+1)) =lim_(n-gtoo) (n/n)/(sqrt(n^2+1)/sqrt(n^2))`

                      `=lim_(n-gtoo) 1/sqrt(1+1/n^2)`

                      `=1/sqrt(1+1/oo)`

                      `= 1/sqrt(1+0)`

                      `= 1 /sqrt(1)`

                      `= 1/1`

                      `=1`

 The limit value `c=1` satisfies ` 0ltclt oo` .   

 Apply the p-series test: `sum_(n=1)^oo 1/n^p` is convergent if ` pgt1` and divergent if `plt=1` .

The `sum_(n=1)^oo 1/n^2`  has `p =2` which satisfy `pgt1` since `2gt1` . Then, the series `sum_(n=1)^oo 1/n^2`  is convergent.

Conclusion based from limit comparison test:

With the series `sum_(n=1)^oo 1/n^2 `  convergent, it follows the series `sum_ (n=1)^oo 1/(nsqrt(n^2+1))` is also convergent.

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