How to find the total number of terms and the sum in a not explicit geometric series? [duplicate]

This series looks like it is geometric or sort of, but the rate between them does not seem constant.

$$S_{n}= \frac{1}{3\times 6}+\frac{1}{6\times 9}+\frac{1}{9\times 12}+ \cdots + \frac{1}{300\times 303}$$

All i can see is the second number in the denominator jumps to the next term in the series but this situation makes it impossible to establish a common factor. Therefore, Is it possible to determine the number of terms and the sum using a simple algorithm?.

There is also a side question which I don't know. In the example from above the final term is $\frac{1}{300\times 303}$ but what if the series goes to the infinity. Is this series convergent or not?. How can this be proven?.

Edit:

Although there is a method to approach these kind of situations mentioned here (What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots +\frac{1}{n(n+1)}$), it does not address the fact on how to obtain a recursive formula, as it is part of the question itself. Therefore is there a way on how to reach to that formula in the denominator and solve the problem?. If the method of solving this involves partial fractions I would appreciate somebody could include a revision of this part in the answer so I can understand how it makes a link with the problem from above.

marked as duplicate by Simply Beautiful Art, Donald Splutterwit, Namaste algebra-precalculus 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(); } ); }); }); Oct 1 '17 at 23:04

• Hint: Partial fractions. – Simply Beautiful Art Oct 1 '17 at 22:34
• In the displayed equation there's an $n$ on the left side but no $n$ on the right side. Should that $S_n$ be $S_{100}?$ – bof Oct 1 '17 at 22:37
• Mind I ask why you are using "$\doteq$" instead of simply "$=$"? – Simply Beautiful Art Oct 1 '17 at 22:49
• @SimplyBeautifulArt Sorry, I didn't notice this when used the Latex editor, now its fixed. – Chris Steinbeck Bell Oct 1 '17 at 22:54
• "it does not address the fact on how to obtain a recursive formula" I tried searching for the word "recursive" and only got one pop-up, which was is the quoted line. – Simply Beautiful Art Oct 1 '17 at 22:56

Define $$a_n = \frac{1}{3n*3(n+1)}$$ Then $$S_n=a_1+a_2+...+a_n$$ Note that $$a_n=\frac{1}{9} \frac{1}{n(n+1)}=\frac{1}{9}(\frac{1}{n}-\frac{1}{n+1})$$ It follows that $$S_n=\frac{1}{9}\left( \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{9}\left( 1-\frac{1}{n+1}\right)$$
• Mind if you sharing a bit on how did you obtained $a_{n}$? – Chris Steinbeck Bell Oct 1 '17 at 23:33
• @ChrisSteinbeckBell How do you define the $n$th fraction in your sum? – Simply Beautiful Art Oct 1 '17 at 23:41