Shell Programming and Scripting

Hi All,
I am facing problem running a script which triggers another script in multiple remote servers. my script in my local server looks like below

Code:

ssh server1 "sudo -u uname /usr/local/script/start.sh &2>&1 >/dev/null "
ssh server2 "sudo -u uname /usr/local/script/start.sh &2>&1 >/dev/null "
ssh server3 "sudo -u uname /usr/local/script/start.sh &2>&1 >/dev/null "

When I run this script, it ssh's to server1 and execute start.sh on server1, but control never come back to command prompt until I kill the process (ctrl+c). So only start.sh on server1 gets executed and have to manually start on server2 and server3.

Could some one please help me getting this resolved?
just an fyi- I am not very much familiar with unix

Thanks,
Sain

On first sight, the &2>&1looks very strange - what do you want to achieve therewith? On the other hand, it should not stop your local script...

I copied from a similar script. I even don't know what it does.
I removed it and tried, but it is still the same, scripts on server2/3 never start.
Also, to mention when i start that script on my local, I see a logs written to my local screen, which is very strange for me..

Pls post the log.

Actual start.sh on remote server starts few jvms which run in background, and logs of jvms starting up and directed to a log directory.
The same start up logs are written on the local screen until i kill that session(but it still logs on remote server, which means those jvms are up and running which is good)
I am trying to figure out how to ssh to a server, execute start.sh, exit that session and go to next remote server and so on.
Do I have to give an exit command after each line?

Pls post script and log.

I found a solution for this from other forums
session would never exit as logs run in back ground, directed logs to /dev/null which helped exiting the session.
Thanks