# How can I prove this series converges? [duplicate]

Wolfram Alpha says the following series converges, but I can't figure out how to prove it.

$$\sum_{n=1}^{\infty}\frac{\sin(n)+\sqrt{n}}{n^2+5}$$

Can I use a comparison test with a simple harmonic p series, or is there a better way?

## marked as duplicate by Mark Viola sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 19 '17 at 17:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• That is quite trivial, my friend: given that $\sum_{n\geq 1}\frac{2\sqrt{n}}{n^2}$ is absolutely convergent, your series is absolutely convergent too. – Jack D'Aurizio Mar 19 '17 at 17:50
• Thanks for whoever found the duplicate. I must be in the same class as this guy... – user427232 Mar 19 '17 at 17:54
• You're welcome! Just FYI ... You can see the person who marked this as a duplicate. – Mark Viola Mar 19 '17 at 18:05

## 2 Answers

Comparison: $$\left|\frac{\sin n + \sqrt n}{n^2+5}\right| \le \frac{2\sqrt n}{n^2} = \frac 2 {n^{3/2}}$$

Hint. If a series is absolutely convergent then it is convergent, observe that, for $n\ge1$, $$\left|\frac{\sin(n)+\sqrt{n}}{n^2+5} \right|\le\frac{\left|\sin(n)+\sqrt{n}\right|}{n^2+5} \le \frac{\left|2\sqrt{n}\right|}{n^2} =2\frac1{n^{3/2}}.$$